5.7: Exponential and Logarithmic Equations
- Use like bases to solve exponential equations.
- Use logarithms to solve exponential equations.
- Use the definition of a logarithm to solve logarithmic equations.
- Use the one-to-one property of logarithms to solve logarithmic equations.
- Solve applied problems involving exponential and logarithmic equations.
In 1859, an Australian landowner named Thomas Austin released \(24\) rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.
Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential and logarithmic equations.
Exponential Equations
Exponential equations are equations that contain variables in the exponent.
Some examples of exponential equations: \[2^x=2^{3x-7} \qquad \qquad 8^{t-3}+4=7\nonumber\]
There are two main techniques to solving exponential equations. The first technique is for solving equations that involve like bases. The second technique is for solving equations that we are unable to write with the same base.
Using Like Bases to Solve Exponential Equations
Consider the equation \(3^{4x−7}=3^{2x}\). Notice that the bases are the same. To have equivalency, the exponents must be equal: \(4x-7=2x\). Solving this, we get \(x=\dfrac{7}{2}\).
The one-to-one property of exponential functions tells us that, for any real numbers \(b\), \(S\), and \(T\), where \(b>0\), \(b≠1\),
\(b^S=b^T\) if and only if \(S=T\).
Solve \(2^{x−1}=2^{2x−4}\)
Solution
\[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]
Solve \(5^{2x}=5^{3x+2}\)
- Answer
-
\(x=−2\)
Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.
For example, consider the equation \(256=4^{x−5}\). We can rewrite both sides of this equation as a power of \(2\). Then we apply the rules of exponents, along with the one-to-one property, to solve for \(x\):
\[\begin{align*} 16&= 2^{x-5}\\ 2^4&= 2^{x-5} \qquad &&\text{Rewrite 16 as a power with base 2}\\ 4&= x-5 \qquad &&\text{Apply the one-to-one property of exponents}\\ 9&= x \qquad &&\text{Add 5 to both sides} \end{align*}\]
Solve \(8^{x+2}={16}^{x+1}\)
Solution
\[\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for x} \end{align*}\]
Solve \(5^{2x}={25}^{3x+2}\)
- Answer
-
\(x=−1\)
Solve \(2^{5x}=\sqrt{2}\)
Solution
\[\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for x} \end{align*}\]
Solve \(5^x=\sqrt{5}\)
- Answer
-
\(x=\dfrac{1}{2}\)
Solving Exponential Equations Using Logarithms
Sometimes the terms of an exponential equation cannot be rewritten with a common base. The one-to-one property for logarithms tells us that, for real numbers \(a>0\) and \(b>0\), \(\log(a)=\log(b)\) is equivalent to \(a=b\). This means that we may apply logarithms with the same base on both sides of an exponential equation. It is most common to apply the natural logarithm on both sides but any logarithms with the same base applied to both sides may be used.
Solve \(5^{x+2}=4^x\)
Solution
\[\begin{align*}
5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\[5pt]
\ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\[5pt]
(x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\[5pt]
x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\[5pt]
x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\[5pt]
x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\[5pt]
x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\[5pt]
x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x}\\[5pt]
x&=\dfrac{-\ln25}{\ln{5}-\ln{4}} \qquad &&\text{simplify using Laws of Logarithms}\end{align*}\]
Solve \(2^x=3^x\)
Solution
\[\begin{align*} \ln2^x&= \ln3^x \qquad &&\text{Take ln of both sides}\\[4pt] x \ln2&= x \ln3 \qquad &&\text{move exponents to the front}\\[4pt] x \ln2-x\ln3&= 0 \qquad &&\text{get like terms on one side}\\[4pt] x(\ln2-\ln3)&= 0 \qquad &&\text{factor out x}\\[4pt] x&=\dfrac{0}{\ln2 - \ln3} \qquad &&\text{divide both sides by} \ln2 - \ln3\\[4pt] x&= 0 \end{align*}\]
Solve \(2^x=3^{x+1}\)
- Answer
-
\(x=\dfrac{\ln3}{\ln2-\ln3 \left (\dfrac{2}{3} \right )}\) or \(x=\dfrac{\ln3}{\ln2-\ln3 }\)
Solve \(5^{x}=11\). Find the exact answer and then approximate it to three decimal places.
Solution :
\(5^{x}=11\)
Since the exponential is isolated, take the logarithm of both sides.
\(\log 5^{x}=\log 11\)
Use the Power Property to get the \(x\) as a factor, not an exponent.
\(x \log 5=\log 11\)
Solve for \(x\). Find the exact answer.
\(x=\frac{\log 11}{\log 5}\)
Approximate the answer.
\(x \approx 1.490\)
Since \(5^{1}=5\) and \(5^{2}=25\), it makes sense that \(5^{1.490}≈11\).
Note in the previous example that \(y=5^x\) is a basic exponential function. We know what the graph looks like and how the function values will start to become large quite quickly for larger x-values since this is an increasing exponential function of base 5, it seems reasonable that the function value of 11 would occur at a smaller x-value. Also, knowing that there is are points on the graph at \((0,1)\) and at \((1,5)\), this would indicate that the function would equal 11 for an x-value that is is just a bit bigger than 1.
Solve \(7^{x}=43\). Find the exact answer and then approximate it to three decimal places.
- Answer
-
\(x=\frac{\log 43}{\log 7} \approx 1.933\)
One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.
Solve \(100=20e^{2t}\)
Solution
\[\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) &&\text{ and } e^x &&\text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}\]
Analysis
Using laws of logs, we can also write this answer in the form \(t=\ln\sqrt{5}\). If we want a decimal approximation of the answer, we use a calculator.
Solve \(3e^{0.5t}=11\)
- Answer
-
\(t=2\ln \left (\dfrac{11}{3} \right )\) or \(\ln{ \left (\dfrac{11}{3} \right )}^2\)
Solve \(4e^{2x}+5=12\)
Solution
\[\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides}\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}\]
Solve \(3+e^{2t}=7e^{2t}\)
- Answer
-
\(t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)\)
Solve \(3e^{x+2}=24\). Find the exact answer and then approximate it to three decimal places.
Solution :
\(3 e^{x+2}=24\)
Isolate the exponential by dividing both sides by \(3\).
\(e^{x+2}=8\)
Take the natural logarithm of both sides.
\(\ln e^{x+2}=\ln 8\)
Use the Power Property to get the \(x\) as a factor, not an exponent.
\((x+2) \ln e=\ln 8\)
Use the property \(\ln e=1\) to simplify.
\(x+2=\ln 8\)
Solve the equation. Find the exact answer.
\(x=\ln 8-2\)
Approximate the answer.
\(x \approx 0.079\)
Solve \(5e^{2x}=25\). Find the exact answer and then approximate it to three decimal places.
- Answer
-
\(x=\frac{\ln 5}{2} \approx 0.805\)
You may be wondering if all exponential equations have solutions. The answer is not all exponential equations have solutions.
The equation \(3^{x+1}=−2\) has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.
Figure \(\PageIndex{2}\) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.
Another example of an equation that has no solution is \(2=−3e^t\).
Figure \(\PageIndex{3}\) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.
Checking for Extraneous Solutions When Solving Exponential Equations
Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.
Solve \(e^{2x}−e^x=56\)
Solution
\[\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero since this equation is of quadratic form}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\end{align*}\]
Solving for each factor, \[ \begin{align*}e^x+7= 0 \qquad \text{or} \qquad e^x-8= 0 \\
e^x= -7 \qquad \text{or} \qquad e^x= 8 \end{align*} \]\(e^x=-7\) has no solution since \(x=\ln(-7)\) is undefined. Solving \(e^x= 8\),\[x= \ln8\nonumber\]
Solve \(e^{2x}=e^x+2\)
- Answer
-
\(x=\ln2\)
Logarithmic Equations
Logarithmic equations are equations that contain logarithms with variables.
Some examples of logarithmic equations: \[\log_5x=8 \qquad \qquad \ln(t)-\ln(t-2)=\ln11\nonumber\]
There are two main techniques to solving logarithmic equations. The first technique is for solving equations that contain only logarithmic terms with the same base. The second technique is for solving equations that contain logarithmic terms and constants.
Using the One-to-One Property of Logarithms to Solve Logarithmic Equations
As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b≠1\),
\({\log}_bS={\log}_bT\) if and only if \(S=T\).
For example,
If \({\log}_2(x−1)={\log}_2(8)\), then \(x−1=8\).
So, if \(x−1=8\), then we can solve for \(x\),and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \({\log}_2(9−1)={\log}_2(8)=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.
For example, consider the equation \(\log(3x−2)−\log(2)=\log(x+4)\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x\):
\[\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\[4pt] \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad &&\text{Apply the quotient rule of logarithms}\\[4pt] \dfrac{3x-2}{2}&= x+4 \qquad &&\text{Apply the one to one property of a logarithm}\\[4pt] 3x-2&= 2x+8 \qquad &&\text{Multiply both sides of the equation by 2}\\[4pt] x&= 10 \qquad &&\text{Subtract 2x and add 2} \end{align*}\]
To check the result, substitute \(x=10\) into \(\log(3x−2)−\log(2)=\log(x+4)\).
\[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\[4pt] \log(28)-\log(2)&= \log(14)\\[4pt] \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad &&\text{The solution checks} \end{align*}\]
Solve \(\ln(x^2)=\ln(2x+3)\)
Solution
\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad &&\text{Use the one-to-one property of the logarithm}\\[4pt] x^2-2x-3&= 0 \qquad &&\text{Get zero on one side before factoring}\\[4pt] (x-3)(x+1)&= 0 \qquad &&\text{Factor}\end{align*}\]
Solving for each factor, \[ x-3= 0 \qquad \text{ or } \qquad x+1=0 \nonumber\] \[x=3 \qquad \text{ or } \qquad x= -1 \nonumber\]
Analysis
There are two solutions: \(3\) or \(−1\). The solution \(−1\) is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.
Solve \(\ln(x^2)=\ln1\)
- Answer
-
\(x=1\) or \(x=−1\)
Using the Definition of a Logarithm to Solve Logarithmic Equations
We have already seen that every logarithmic equation \({\log}_b(x)=y\) is equivalent to the exponential equation \(b^y=x\). We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.
For example, consider the equation \({\log}_2(2)+{\log}_2(3x−5)=3\). To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):
\[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\[4pt] {\log}_2(2(3x-5))&= 3 \qquad &&\text{Apply the product rule of logarithms}\\[4pt] {\log}_2(6x-10)&= 3 \qquad &&\text{Distribute}\\[4pt] 2^3&= 6x-10 \qquad &&\text{Convert to exponential form}\\[4pt] 8&= 6x-10 \qquad &&\text{Calculate } 2^3\\[4pt] 18&= 6x \qquad &&\text{Add 10 to both sides}\\[4pt] x&= 3 \qquad &&\text{Divide by 6} \end{align*}\]
Solve \(2\ln x+3=7\)
Solution
\[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad &&\text{Subtract 3}\\ \ln x&= 2 \qquad &&\text{Divide by 2}\\ x&= e^2 \qquad &&\text{Rewrite in exponential form} \end{align*}\]
Solve \(6+\ln x=10\)
- Answer
-
\(x=e^4\)
Solve \(2\ln(6x)=7\)
Solution
\[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad &&\text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad &&\text{convert to exponential form}\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad &&\text{Divide by 6} \end{align*}\]
Solve \(2\ln(x+1)=10\)
- Answer
-
\(x=e^5−1\)
Solve \(\ln x=3\)
Solution
\[\begin{align*} \ln x&= 3\\ x&= e^3 \end{align*}\]
Figure \(\PageIndex{4}\) represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to \(20\). In other words \(e^3≈20\). A calculator gives a better approximation: \(e^3≈20.0855\).
Use a graphing calculator to estimate the approximate solution to the logarithmic equation \(2^x=1000\) to \(2\) decimal places.
- Answer
-
\(x≈9.97\)
Checking for Extraneous Solutions When Solving Logarithmic Equations
It is possible to have extraneous solutions when solving logarithmic equations since the domain of a logarithm is \((0,\infty)\). Check your answers to see if they are extraneous or not.
Solve \(\ln{x}+\ln(2x-5) = \ln{3}\)
Solution
\(\begin{align*} \ln{x}+\ln(2x-5) &= \ln3 \\[4pt] \ln(x(2x-5)) &= \ln3 \\[4pt] \ln(2x^2-5x) &= \ln3 \\[4pt] 2x^2-5x &= 3 \\[4pt] 2x^2-5x-3 &= 0 \\[4pt] (2x+1)(x-3) &=0 \\[4pt] 2x+1=0 \qquad x-3&=0 \\[4pt] x=\dfrac{1}{2} \qquad x&=3 \\[4pt] \end{align*}\)
Solve: \(\log _{4}(x+6)-\log _{4}(2 x+5)=-\log _{4} x\)
Solution :
\(\log _{4}(x+6)-\log _{4}(2 x+5)=-\log _{4} x\)
Use the Quotient Property on the left side and the Power Property on the right.
\(\log _{4}\left(\dfrac{x+6}{2 x+5}\right)=\log _{4} x^{-1}\)
Rewrite \(x^{-1}=\dfrac{1}{x}\).
\(\log _{4}\left(\dfrac{x+6}{2 x+5}\right)=\log _{4} \dfrac{1}{x}\)
Use the One-to-One Property, if \(\log _{a} M=\log _{a} N\), then \(M=N\).
\(\dfrac{x+6}{2 x+5}=\dfrac{1}{x}\)
Solve the rational equation.
\(x(x+6)=2 x+5\)
Distribute.
\(x^{2}+6 x=2 x+5\)
Write in standard form.
\(x^{2}+4 x-5=0\)
Factor.
\((x+5)(x-1)=0\)
Use the Zero-Product-Property.
\(x+5=0, \quad x-1=0\)
Solve each equation.
\(\cancel{x=-5}, \quad x=1\)
Check.
We leave the check for you.
Solve: \(2 \log _{5} x=\log _{5} 81\)
Solution :
\(2 \log _{5} x=\log _{5} 81\)
Use the Power Property.
\(\log _{5} x^{2}=\log _{5} 81\)
Use the One-to-One Property, if \(\log _{a} M=\log _{a} N\), then \(M=N\).
\(x^{2}=81\)
Solve using the Square Root Property.
\(x=\pm 9\)
We eliminate \(x=-9\) as we cannot take the logarithm of a negative number.
\(x=9, \cancel{x=-9}\)
Check. \(x=9\)
\(\begin{aligned}2 \log _{5} x&=\log _{5} 81 \\ 2 \log _{5} 9 &\stackrel{?}{=} \log _{5} 81 \\ \log _{5} 9^{2} & \stackrel{?}{=}\log _{5} 81 \\ \log _{5} 81 & =\log _{5} 81\end{aligned}\)
Solve: \(2 \log _{3} x=\log _{3} 36\)
- Answer
-
\(x=6\)
Solve: \(\log (x+2)-\log (4 x+3)=-\log x\)
- Answer
-
\(x=3\)
Solve: \(\log (x-2)-\log (4 x+16)=\log \dfrac{1}{x}\)
- Answer
-
\(x=8\)
Solving Applied Problems
In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.
Suppose \($2000\) is invested in an account which offers \(7.125 \%\) compounded monthly.
- Express the amount \(A\) in the account as a function of the term of the investment \(t\) in years.
- How much is in the account after \(5\) years?
- How long will it take for the initial investment to double?
- Find and interpret the average rate of change of the amount in the account from the end of the fourth year to the end of the fifth year, and from the end of the thirty-fourth year to the end of the thirty-fifth year.
Solution
- Substituting \(P = 2000\), \(r = 0.07125\), and \(n = 12\) (since interest is compounded monthly into the compound interest formula yields \[A(t) = 2000\left(1 + \dfrac{0.07125}{12}\right)^{12t}=2000 (1.0059375)^{12t}\nonumber\]
- Since \(t\) represents the length of the investment in years, we substitute \(t=5\) into \(A(t)\) to find \[A(5) = 2000 (1.0059375)^{12(5)} \approx 2852.92\nonumber\]After \(5\) years, we have approximately \($2852.92\).
- Our initial investment is \($2000\), so to find the time it takes this to double, we need to find \(t\) when \(A(t) = 4000\). We get \(2000 (1.0059375)^{12t}=4000\), or \((1.0059375)^{12t}=2\). Taking natural logs of both sides, we get \[t = \dfrac{\ln(2)}{12 \ln(1.0059375)} \approx 9.75.\nonumber\] Hence, it takes approximately \(9\) years \(9\) months for the investment to double.
- To find the average rate of change of \(A\) from the end of the fourth year to the end of the fifth year, we compute \[\frac{A(5)-A(4)}{5-4} \approx 195.63\nonumber\] Similarly, the average rate of change of \(A\) from the end of the thirty-fourth year to the end of the thirty-fifth year is \[\frac{A(35)-A(34)}{35-34} \approx 1648.21\nonumber\] This means that the value of the investment is increasing at a rate of approximately \($195.63\) per year between the end of the fourth and fifth years, while that rate jumps to \($1648.21\) per year between the end of the thirty-fourth and thirty-fifth years. So, not only is it true that the longer you wait, the more money you have, but also the longer you wait, the faster the money increases. (in fact, the rate of increase of the amount in the account is exponential as well. This is the quality that really defines exponential functions and we refer the reader to a course in Calculus.


17.3: Rate Laws
- Explain the form and function of a rate law
- Use rate laws to calculate reaction rates
- Use rate and concentration data to identify reaction orders and derive rate laws
As described in the previous module, the rate of a reaction is affected by the concentrations of reactants. Rate laws or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. In general, a rate law (or differential rate law, as it is sometimes called) takes this form:
\[\ce{rate}=k[A]^m[B]^n[C]^p… \nonumber \]
in which [ A ], [ B ], and [ C ] represent the molar concentrations of reactants, and k is the rate constant , which is specific for a particular reaction at a particular temperature. The exponents m , n , and p are usually positive integers (although it is possible for them to be fractions or negative numbers). The rate constant k and the exponents m , n , and p must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the concentration of A , B , or C , but it does vary with temperature and surface area.
The exponents in a rate law describe the effects of the reactant concentrations on the reaction rate and define the reaction order . Consider a reaction for which the rate law is:
\[\ce{rate}=k[A]^m[B]^n \nonumber \]
If the exponent m is 1, the reaction is first order with respect to A . If m is 2, the reaction is second order with respect to A . If n is 1, the reaction is first order in B . If n is 2, the reaction is second order in B . If m or n is zero, the reaction is zero order in A or B , respectively, and the rate of the reaction is not affected by the concentration of that reactant. The overall reaction order is the sum of the orders with respect to each reactant. If m = 1 and n = 1, the overall order of the reaction is second order ( m + n = 1 + 1 = 2).
The rate law:
\[\ce{rate}=k[\ce{H2O2}] \nonumber \]
describes a reaction that is first order in hydrogen peroxide and first order overall. The rate law:
\[\ce{rate}=k[\ce{C4H6}]^2 \nonumber \]
describes a reaction that is second order in C 4 H 6 and second order overall. The rate law:
\[\ce{rate}=k[\ce{H+}][\ce{OH-}] \nonumber \]
describes a reaction that is first order in H + , first order in OH − , and second order overall.
An experiment shows that the reaction of nitrogen dioxide with carbon monoxide:
\[\ce{NO2}(g)+\ce{CO}(g)⟶\ce{NO}(g)+\ce{CO2}(g) \nonumber \]
is second order in NO 2 and zero order in CO at 100 °C. What is the rate law for the reaction?
Solution
The reaction will have the form:
\[\ce{rate}=k[\ce{NO2}]^m[\ce{CO}]^n \nonumber \]
The reaction is second order in NO 2 ; thus m = 2. The reaction is zero order in CO; thus n = 0. The rate law is:
\[\ce{rate}=k[\ce{NO2}]^2[\ce{CO}]^0=k[\ce{NO2}]^2 \nonumber \]
Remember that a number raised to the zero power is equal to 1, thus [CO] 0 = 1, which is why we can simply drop the concentration of CO from the rate equation: the rate of reaction is solely dependent on the concentration of NO 2 . When we consider rate mechanisms later in this chapter, we will explain how a reactant’s concentration can have no effect on a reaction despite being involved in the reaction.
The rate law for the reaction:
\[\ce{H2}(g)+\ce{2NO}(g)⟶\ce{N2O}(g)+\ce{H2O}(g) \nonumber \]
has been experimentally determined to be rate = k [NO] 2 [H 2 ]. What are the orders with respect to each reactant, and what is the overall order of the reaction?
- Answer
-
- order in NO = 2;
- order in H 2 = 1;
- overall order = 3
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH 3 OH) and ethyl acetate (CH 3 CH 2 OCOCH 3 ) as a sample reaction before studying the chemical reactions that produce biodiesel:
\[\ce{CH3OH + CH3CH2OCOCH3 ⟶ CH3OCOCH3 + CH3CH2OH} \nonumber \]
The rate law for the reaction between methanol and ethyl acetate is, under certain conditions, experimentally determined to be:
\[\ce{rate}=k[\ce{CH3OH}] \nonumber \]
What is the order of reaction with respect to methanol and ethyl acetate, and what is the overall order of reaction?
- Answer
-
- order in CH 3 OH = 1;
- order in CH 3 CH 2 OCOCH 3 = 0;
- overall order = 1
It is sometimes helpful to use a more explicit algebraic method, often referred to as the method of initial rates , to determine the orders in rate laws. To use this method, we select two sets of rate data that differ in the concentration of only one reactant and set up a ratio of the two rates and the two rate laws. After canceling terms that are equal, we are left with an equation that contains only one unknown, the coefficient of the concentration that varies. We then solve this equation for the coefficient.
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (Figure \(\PageIndex{1}\)). One such reaction is the combination of nitric oxide, NO, with ozone, O 3 :
\[\ce{NO}(g)+\ce{O3}(g)⟶\ce{NO2}(g)+\ce{O2}(g) \nonumber \]
This reaction has been studied in the laboratory, and the following rate data were determined at 25 °C.
| Trial | \([\ce{NO}]\) (mol/L) | \([\ce{O3}]\) (mol/L) | \(\dfrac{Δ[\ce{NO2}]}{Δt}\:\mathrm{(mol\:L^{−1}\:s^{−1})}\) |
|---|---|---|---|
| 1 | 1.00 × 10 −6 | 3.00 × 10 −6 | 6.60 × 10 −5 |
| 2 | 1.00 × 10 −6 | 6.00 × 10 −6 | 1.32 × 10 −4 |
| 3 | 1.00 × 10 −6 | 9.00 × 10 −6 | 1.98 × 10 −4 |
| 4 | 2.00 × 10 −6 | 9.00 × 10 −6 | 3.96 × 10 −4 |
| 5 | 3.00 × 10 −6 | 9.00 × 10 −6 | 5.94 × 10 −4 |
Determine the rate law and the rate constant for the reaction at 25 °C.
Solution
The rate law will have the form:
\[\ce{rate}=k[\ce{NO}]^m[\ce{O3}]^n \nonumber \]
We can determine the values of m , n , and k from the experimental data using the following three-part process:
- Determine the value of m from the data in which [NO] varies and [O 3 ] is constant. In the last three experiments, [NO] varies while [O 3 ] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
-
Determine the value of
n
from data in which [O
3
] varies and [NO] is constant.
In the first three experiments, [NO] is constant and [O
3
] varies. The reaction rate changes in direct proportion to the change in [O
3
]. When [O
3
] doubles from trial 1 to 2, the rate doubles; when [O
3
] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O
3
], and
n
is equal to 1.The rate law is thus:
\[\ce{rate}=k[\ce{NO}]^1[\ce{O3}]^1=k[\ce{NO}][\ce{O3}] \nonumber \]
-
Determine the value of
k
from one set of concentrations and the corresponding rate
.
\[\begin{align*}
k&=\mathrm{\dfrac{rate}{[NO][O_3]}}\\
&=\mathrm{\dfrac{6.60×10^{−5}\cancel{mol\: L^{−1}}\:s^{−1}}{(1.00×10^{−6}\cancel{mol\: L^{−1}})(3.00×10^{−6}\:mol\:L^{−1})}}\\
&=\mathrm{2.20×10^7\:L\:mol^{−1}\:s^{−1}}
\end{align*} \nonumber \]The large value of k tells us that this is a fast reaction that could play an important role in ozone depletion if [NO] is large enough.
Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation:
\[\ce{CH3CHO}(g)⟶\ce{CH4}(g)+\ce{CO}(g) \nonumber \]
Determine the rate law and the rate constant for the reaction from the following experimental data:
| Trial | \([\ce{CH3CHO}]\) (mol/L) | \(−\dfrac{Δ[\ce{CH3CHO}]}{Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}\) |
|---|---|---|
| 1 | 1.75 × 10 −3 | 2.06 × 10 −11 |
| 2 | 3.50 × 10 −3 | 8.24 × 10 −11 |
| 3 | 7.00 × 10 −3 | 3.30 × 10 −10 |
- Answer
-
\(\ce{rate}=k[\ce{CH3CHO}]^2\) with k = 6.73 × 10 −6 L/mol/s
Using the initial rates method and the experimental data, determine the rate law and the value of the rate constant for this reaction:
\[\ce{2NO}(g)+\ce{Cl2}(g)⟶\ce{2NOCl}(g) \nonumber \]
| Trial | [NO] (mol/L) | \([Cl_2]\) (mol/L) | \(−\dfrac{Δ[\ce{NO}]}{2Δt}\mathrm{(mol\:L^{−1}\:s^{−1})}\) |
|---|---|---|---|
| 1 | 0.10 | 0.10 | 0.00300 |
| 2 | 0.10 | 0.15 | 0.00450 |
| 3 | 0.15 | 0.10 | 0.00675 |
Solution
The rate law for this reaction will have the form:
\[\ce{rate}=k[\ce{NO}]^m[\ce{Cl2}]^n \nonumber \]
As in Example \(\PageIndex{2}\), we can approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k . In this example, however, we will use a different approach to determine the values of m and n :
Determine the value of m from the data in which [NO] varies and [Cl 2 ] is constant. We can write the ratios with the subscripts x and y to indicate data from two different trials:
\[\dfrac{\ce{rate}_x}{\ce{rate}_y}=\dfrac{k[\ce{NO}]^m_x[\ce{Cl2}]^n_x}{k[\ce{NO}]^m_y[\ce{Cl2}]^n_y} \nonumber \]
Using the third trial and the first trial, in which [Cl 2 ] does not vary, gives:
\[\mathrm{\dfrac{rate\: 3}{rate\: 1}}=\dfrac{0.00675}{0.00300}=\dfrac{k(0.15)^m(0.10)^n}{k(0.10)^m(0.10)^n} \nonumber \]
After canceling equivalent terms in the numerator and denominator, we are left with:
\[\dfrac{0.00675}{0.00300}=\dfrac{(0.15)^m}{(0.10)^m} \nonumber \]which simplifies to:
\[2.25=(1.5)^m \nonumber \]
We can use natural logs to determine the value of the exponent m :
\ln(2.25)&=m\ln(1.5)
\dfrac{\ln(2.25)}{\ln(1.5)}&=m
2&=m
\end{align*}\)
We can confirm the result easily, since:
-
Determine the value of
n
from data in which [Cl
2
] varies and [NO] is constant
. \[\mathrm{\dfrac{rate\: 2}{rate\: 1}}=\dfrac{0.00450}{0.00300}=\dfrac{k(0.10)^m(0.15)^n}{k(0.10)^m(0.10)^n} \nonumber \]
Cancelation gives:
\[\dfrac{0.0045}{0.0030}=\dfrac{(0.15)^n}{(0.10)^n} \nonumber \]
which simplifies to:
\[1.5=(1.5)^n \nonumber \]
Thus n must be 1, and the form of the rate law is:
\[\ce{Rate}=k[\ce{NO}]^m[\ce{Cl2}]^n=k[\ce{NO}]^2[\ce{Cl2}] \nonumber \]
-
Determine the numerical value of the rate constant
k
with appropriate units.
The units for the rate of a reaction are mol/L/s. The units for
k
are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol
3
/L
3
. The units for
k
should be mol
−2
L
2
/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k :
\(\begin{align*}
\mathrm{0.00300\:mol\:L^{−1}\:s^{−1}}&=k\mathrm{(0.10\:mol\:L^{−1})^2(0.10\:mol\:L^{−1})^1}\\
k&=\mathrm{3.0\:mol^{−2}\:L^2\:s^{−1}}
\end{align*}\)
Use the provided initial rate data to derive the rate law for the reaction whose equation is:
\[\ce{OCl-}(aq)+\ce{I-}(aq)⟶\ce{OI-}(aq)+\ce{Cl-}(aq) \nonumber \]
| Trial | [OCl − ] (mol/L) | [I − ] (mol/L) | Initial Rate (mol/L/s) |
|---|---|---|---|
| 1 | 0.0040 | 0.0020 | 0.00184 |
| 2 | 0.0020 | 0.0040 | 0.00092 |
| 3 | 0.0020 | 0.0020 | 0.00046 |
Determine the rate law expression and the value of the rate constant k with appropriate units for this reaction.
- Answer
-
\(\mathrm{\dfrac{rate\: 2}{rate\: 3}}=\dfrac{0.00092}{0.00046}=\dfrac{k(0.0020)^x(0.0040)^y}{k(0.0020)^x(0.0020)^y}\)
2.00 = 2.00 y
y = 1
\(\mathrm{\dfrac{rate\: 1}{rate\: 2}}=\dfrac{0.00184}{0.00092}=\dfrac{k(0.0040)^x(0.0020)^y}{k(0.0020)^x(0.0040)^y}\)
\(\begin{align*}
2.00&=\dfrac{2^x}{2^y}\\
2.00&=\dfrac{2^x}{2^1}\\
4.00&=2^x\\
x&=2
\end{align*}\)
Substituting the concentration data from trial 1 and solving for k yields:
\(\begin{align*}
\ce{rate}&=k[\ce{OCl-}]^2[\ce{I-}]^1\\
0.00184&=k(0.0040)^2(0.0020)^1\\
k&=\mathrm{5.75×10^4\:mol^{−2}\:L^2\:s^{−1}}
\end{align*}\)
Reaction Order and Rate Constant Units
In some of our examples, the reaction orders in the rate law happen to be the same as the coefficients in the chemical equation for the reaction. This is merely a coincidence and very often not the case. Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
\(\ce{NO2 + CO⟶NO + CO2}\hspace{20px}\ce{rate}=k[\ce{NO2}]^2\\
\ce{CH3CHO⟶CH4 + CO}\hspace{20px}\ce{rate}=k[\ce{CH3CHO}]^2\\
\ce{2N2O5⟶2NO2 + O2}\hspace{20px}\ce{rate}=k[\ce{N2O5}]\\
\ce{2NO2 + F2⟶2NO2F}\hspace{20px}\ce{rate}=k[\ce{NO2}][\ce{F2}]\\
\ce{2NO2Cl⟶2NO2 + Cl2}\hspace{20px}\ce{rate}=k[\ce{NO2Cl}]\)
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
Reaction orders also play a role in determining the units for the rate constant k . In Example \(\PageIndex{2}\), a second-order reaction, we found the units for k to be \(\mathrm{L\:mol^{-1}\:s^{-1}}\), whereas in Example \(\PageIndex{3}\), a third order reaction, we found the units for k to be mol −2 L 2 /s. More generally speaking, the units for the rate constant for a reaction of order \( (m+n)\) are \(\ce{mol}^{1−(m+n)}\ce L^{(m+n)−1}\ce s^{−1}\). Table \(\PageIndex{1}\) summarizes the rate constant units for common reaction orders.
| Reaction Order | Units of k |
|---|---|
| \( (m+n)\) | \(\ce{mol}^{1−(m+n)}\ce L^{(m+n)−1}\ce s^{−1}\) |
| zero | mol/L/s |
| first | s −1 |
| second | L/mol/s |
| third | mol −2 L 2 s −1 |
Note that the units in the table can also be expressed in terms of molarity ( M ) instead of mol/L. Also, units of time other than the second (such as minutes, hours, days) may be used, depending on the situation.
Summary
Rate laws provide a mathematical description of how changes in the amount of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the amount of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction.
Glossary
- method of initial rates
- use of a more explicit algebraic method to determine the orders in a rate law
- overall reaction order
- sum of the reaction orders for each substance represented in the rate law
- rate constant ( k )
- proportionality constant in the relationship between reaction rate and concentrations of reactants
- rate law
- (also, rate equation) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
- reaction order
- value of an exponent in a rate law, expressed as an ordinal number (for example, zero order for 0, first order for 1, second order for 2, and so on)


6.11: Python Examples
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Overview
The following examples demonstrate data types, arithmetic operations, and input in Python.
Data Types
# This program demonstrates variables, literal constants, and data types. i =<PHONE_NUMBER> f = 1.23456789012345 s = "string" b = True print("Integer i =", i) print("Float f =", f) print("String s =", s) print("Boolean b =", b)
Output
Integer i =<PHONE_NUMBER> Float f = 1.23456789012345 String s = string Boolean b = true
Discussion
Each code element represents:
-
#
begins a comment -
i = , d = , s =, b =
assign literal values to the corresponding variables -
print()
calls the print function
Arithmetic
# This program demonstrates arithmetic operations. a = 3 b = 2 print("a =", a) print("b =", b) print("a + b =", (a + b)) print("a - b =", (a - b)) print("a * b =", a * b) print("a / b =", a / b) print("a % b =", (a % b))
Output
a = 3 b = 2 a + b = 5 a - b = 1 a * b = 6 a / b = 1.5 a % b = 1
Discussion
Each new code element represents:
-
+, -, *, /, and %
represent addition, subtraction, multiplication, division, and modulus, respectively.
Temperature
# This program converts an input Fahrenheit temperature to Celsius. print("Enter Fahrenheit temperature:") fahrenheit = float(input()) celsius = (fahrenheit - 32) * 5 / 9 print(str(fahrenheit) + "° Fahrenheit is " + str(celsius) + "° Celsius")
Output
Enter Fahrenheit temperature: 100 100.0° Fahrenheit is 37.77777777777778° Celsius
Discussion
Each new code element represents:
-
input()
reads the next line from standard input -
float()
converts the input to a floating-point value
References
- Wikiversity: Computer Programming


11.1: What Is a Fluid?
Learning Objectives
By the end of this section, you will be able to:
- State the common phases of matter.
- Explain the physical characteristics of solids, liquids, and gases.
- Describe the arrangement of atoms in solids, liquids, and gases.
Matter most commonly exists as a solid, liquid, or gas; these states are known as the three common phases of matter . Solids have a definite shape and a specific volume, liquids have a definite volume but their shape changes depending on the container in which they are held, and gases have neither a definite shape nor a specific volume as their molecules move to fill the container in which they are held (Figure \(\PageIndex{1}\)). Liquids and gases are considered to be fluids because they yield to shearing forces, whereas solids resist them. Note that the extent to which fluids yield to shearing forces (and hence flow easily and quickly) depends on a quantity called the viscosity which is discussed in detail in Viscosity and Laminar Flow; Poiseuille’s Law . We can understand the phases of matter and what constitutes a fluid by considering the forces between atoms that make up matter in the three phases.
Atoms in solids are in close contact, with forces between them that allow the atoms to vibrate but not to change positions with neighboring atoms. (These forces can be thought of as springs that can be stretched or compressed, but not easily broken.) Thus a solid resists all types of stress. A solid cannot be easily deformed because the atoms that make up the solid are not able to move about freely. Solids also resist compression, because their atoms form part of a lattice structure in which the atoms are a relatively fixed distance apart. Under compression, the atoms would be forced into one another. Most of the examples we have studied so far have involved solid objects which deform very little when stressed.
Submicroscopic Explanation of Solids and Liquids
Atomic and molecular characteristics explain and underlie the macroscopic characteristics of solids and fluids. This submicroscopic explanation is one theme of this text and is highlighted in Conservation of Momentum . See, for example, microscopic description of collisions and momentum or microscopic description of pressure in a gas. This present section is devoted entirely to the submicroscopic explanation of solids and liquids.
- In contrast, liquids deform easily when stressed and do not spring back to their original shape once the force is removed because the atoms are free to slide about and change neighbors—that is, they flow (so they are a type of fluid), with the molecules held together by their mutual attraction. When a liquid is placed in a container with no lid on, it remains in the container (providing the container has no holes below the surface of the liquid!). Because the atoms are closely packed, liquids, like solids, resist compression.
Atoms in gases are separated by distances that are large compared with the size of the atoms. The forces between gas atoms are therefore very weak, except when the atoms collide with one another. Gases thus not only flow (and are therefore considered to be fluids) but they are relatively easy to compress because there is much space and little force between atoms. When placed in an open container gases, unlike liquids, will escape. The major distinction is that gases are easily compressed, whereas liquids are not. We shall generally refer to both gases and liquids simply as fluids , and make a distinction between them only when they behave differently.
PhET Explorations: States of Matter - Basics
Heat, cool, and compress atoms and molecules and watch as they change between solid, liquid, and gas phases.
Summary
- A fluid is a state of matter that yields to sideways or shearing forces. Liquids and gases are both fluids. Fluid statics is the physics of stationary fluids.
Glossary
- fluids
- liquids and gases; a fluid is a state of matter that yields to shearing forces


Skip to main content
-
aldosterone
-
a hormone made in the adrenal cortex that helps to control the balance of water and salts in the kidneys, retaining sodium and releasing potassium from the body
-
angina
-
chest discomfort in the front of the chest, neck, shoulders, jaw, or arms that is precipitated by physical exertions and relieved by rest or sublingual nitrates
-
angiotensin I
-
a protein in blood that promotes aldosterone secretion and raises blood pressure
-
angiotensin II
-
a protein in the blood that causes the muscular walls of the arterioles to constrict and narrow, thereby increasing blood pressure
-
angiotensin II receptor blocker (ARB)
-
a classification of drug that binds to and inhibits angiotensin II type I receptors
-
angiotensin-converting enzyme (ACE) inhibitor
-
a classification of drug that blocks the body’s production of angiotensin II; the protein that causes vasoconstriction and inhibits the reuptake of norepinephrine, which stimulates catecholamine release
-
antianginal drugs
-
drugs used in the treatment of angina
-
antihypertensive drugs
-
drugs used in the treatment of hypertension
-
beta-adrenergic blocker
-
a classification of drug that inhibits chronotropic, inotropic, and vasoconstrictor response to catecholamine, epinephrine, and norepinephrine by exerting effects on adrenergic receptors beta 1, beta 2, and alpha.
-
calcium channel blocker
-
a classification of drug that blocks calcium from entering cells by binding to long-acting voltage-gated calcium channels in the heart, smooth muscle, and pancreas
-
cardiac output
-
the product of the heart rate and stroke volume; the volume of blood pumped by the heart per unit of time, usually measured in liters per minute
-
diastolic blood pressure
-
indicates how much pressure the blood is exerting against artery walls while the heart is resting between beats; the second number of a blood pressure reading
-
diuretic
-
a classification of drug that induces sodium loss and increases urine flow; typically used to treat hypertension, heart failure, and volume overload states
-
heart rate
-
the number of times each minute the heart beats
-
hypertension
-
when an individual’s blood pressure is above the normal limits for a sustained period of time
-
hypotension
-
when an individual’s blood pressure is below the normal limits for a sustained period of time
-
nitrate
-
a classification of drug that causes vasodilation of blood vessels by imparting nitric oxide, which relaxes smooth muscles
-
peripheral vascular resistance
-
determined by blood flow in the body and the level of constriction or dilatation within the vessels
-
renin-angiotensin-aldosterone system (RAAS)
-
a compensatory mechanism the body activates during hypotension (when the blood pressure is low)
-
stroke volume
-
the volume of blood pumped out of the left ventricle of the heart during each systolic cardiac contraction
-
systolic blood pressure
-
indicates how much pressure the blood is exerting against the artery walls when the heart beats; the first number of a blood pressure reading
-
vasopressin
-
an antidiuretic hormone that regulates blood pressure, blood osmolality, and blood volume


3.19: Discussion- Paragraph Styles
Watch the five-minute video below. While the video shows an older version of Word (Microsoft Word 2010), the same features and many of the same buttons are also found in other versions of Word, including 2019 and 365.
A link to an interactive elements can be found at the bottom of this page.
You can also view the transcript for the video “Word 2010: Paragraph Styles” here (opens in new window).
Discussion Questions
- In what situations would the Styles feature be useful?
- Would you ever use this feature?
- Why or why not?
Share your opinions below and respond to two of your classmates’ thoughts.
Contributors and Attributions
CC licensed content, Shared previously
- Word 2010: Paragraph Styles. Authored by : Chris Walker. Located at : https://www.youtube.com/watch?v=7pw9RtCET0w . License : CC BY: Attribution
Lumen Learning authored content
- Discussion: Paragraph Styles. Provided by : Lumen Learning. License : CC BY: Attribution


11.14: Quiz 13-Electrical Safety
1. True or False... The "force" carried by electricity is measured in "volts"?
2. True or False... It is "amps" (the intensity of current) which deliver electric shocks, not volts
3. How many "amps" can most household and industrial lines safely carry? ________
4. True or False.... 0.06 amps (the amount of electricity needed to light a Christmas tree bulb) can be fatal?
5. True or False.... If you are using a "three-prong" outlet, you can be sure that it is"grounded"?
6. True or False.... If one of the fuses in a fuse box is continually burning out, you should replace it with a higher rated fuse?
7. True or False.... If electrical equipment is running when a leak of a flammable gas or vapor develops in your work area, you should immediately turn the equipment off?


2.6: Financial Claims Hierarchy
Lenders and owners have different claims on the company’s interests. There is a distinct hierarchy in which the corporation’s claimants get paid – in order from first to last:
Debtholders : They get paid first. This includes the interest on loans, and on the loan’s principal when due. Loan payments are made as contracted, and do not increase should the firm become more profitable. Dividends may not be paid to shareholders unless loan payments have been made first, and in full.
Preferred shareholders : These owners (usually) get paid a fixed payment or “dividend,” which cannot increase even if the firm’s profits increase. If the firm is unable to pay the dividend it may skip it. However, most preferred dividends are “cumulative,” which means that no common stock dividends may be paid unless and until all past unpaid dividends that have accumulated “in arrears” are paid in full first.
Common shareholders : These shareholders get their dividends last, and are effective owners of any earnings, which the firm does not pay out, but “retains” and reinvests back into the firm in the manner of added property, plant, equipment, and working capital. In other words, common shareholders have a claim on the firm’s net income (after preferred dividends have been paid).
Common shareholders take on the most risk as they are last ones “on the totem pole.” Such is the case whether the firm is an ongoing enterprise or is being liquidated in bankruptcy. They get paid last. Common shareholders have a “residual” claim, or interest, in the company – after all other interests are taken care of.
On the other hand, the common shareholders have the most to gain if the firm is profitable; the dividend may be increased and more earnings may be retained – to their financial benefit, as the firm’s “retained earnings” are owned by the common shareholders.
Again, only common shareholders benefit if profits go up; common share ownership entails more risk, since if interest and preferred dividends are not paid, there may be nothing left for the common shareholders.
Government and Taxes : Let’s not forget that, after interest has been paid, the earnings of the company are then taxed, and that dividends are paid out of Net Income – after taxes have been paid. Taxes are a given, and normally, we do not think of the government as a claimant on the firm, since it is neither lender nor owner.
Nothing is certain except death and taxes.
-Benjamin Franklin


1.9: Sociology Today
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- LibreTexts
Contemporary sociology does not have a single overarching foundation—it has varying methods, both qualitative and quantitative.
Learning Objectives
- Describe how the discipline of sociology has expanded since its foundation
Key Points
- The traditional focuses of sociology have included social stratification, social class, culture, social mobility, religion, secularization, law, and deviance.
- Sociology has gradually expanded its focus to include more diverse subjects such as health, medical, penal institutions, the Internet, or the role of social activity in the development of scientific knowledge.
- The linguistic and cultural turns of the mid-twentieth century led to increasingly interpretative, hermeneutic, and philosophic approaches to the analysis of society.
Key Terms
- secularization : The transformation of a society from close identification with religious values and institutions toward non-religious (or “irreligious”) values and secular institutions.
- hermeneutic : Something that explains, interprets, illustrates or elucidates.
- paradigm : A system of assumptions, concepts, values, and practices that constitutes a way of viewing reality.
Although sociology emerged from Comte’s vision of a discipline that would subsume all other areas of scientific inquiry, that was the future of sociology. Far from replacing the other sciences, contemporary sociology has taken its place as a particular perspective for investigating human social life.
The traditional focuses of sociology have included social stratification, social class, culture, social mobility, religion, secularization, law, and deviance. As all spheres of human activity are affected by the interplay between social structure and individual agency, sociology has gradually expanded to focus on more diverse subjects such as health, medical, military and penal institutions, the Internet, and the role of social activity in the development of scientific knowledge.
The range of social scientific methodology has also expanded. Social researchers draw upon a variety of qualitative and quantitative techniques. The linguistic and cultural turns of the mid-twentieth century led to increasingly interpretative, hermeneutic, and philosophic approaches to the analysis of society. Conversely, recent decades have seen the rise of new analytically, mathematically, and computationally rigorous techniques such as agent-based modelling and social network analysis.
Presently, sociological theories lack a single overarching foundation, and there is little consensus about what such a framework should consist of. However, a number of broad paradigms cover much modern sociological theorizing. In the humanistic parts of the discipline, these paradigms are referred to as social theory, often shared with the humanities. The discipline’s dominant scientifically-oriented areas generally focus on a different set of theoretical perspectives, generally referred to as sociological theory. These include new institutionalism, social networks, social identity, social and cultural capital, toolkit and cognitive theories of culture, and resource mobilization. Analytical sociology is an ongoing effort to systematize many of these middle-range theories.


11.3: Managing and Facilitating Movement Dynamics
Some ecosystems are transient in nature—their character is temporary and will change because of disturbance and succession. Consequently, species that occupy those transient habitats are bound to be naturally extirpated at one time or another. Consider, for example, a small population of wildflowers occurring in a river’s floodplain; at some stage, there is going to be a flood that will wash away those flowers. But the flooding also disperses seeds downstream, allowing for new wildflower populations to establish in suitable habitat elsewhere. These shifting populations linked by movements between them are better characterized as a metapopulation (a “population of populations”) (Figure 11.8) consisting of several subpopulations. For some metapopulations, every subpopulation is transient: their distribution changes dramatically with each generation. Other metapopulations involve relatively permanent subpopulations with only a few individuals dispersing each generation. Some metapopulations consist of one or more source populations whose sizes are stable or increasing, and several sink populations whose sizes fluctuate depending on environmental conditions. Some sink subpopulations may undergo such dramatic fluctuations that they would be extirpated in unfavourable years were it not for population rescue by immigrants from source populations.
A metapopulation (a “population of populations”) consists of several subpopulations linked by movements of individuals between them.
Habitat fragmentation threatens metapopulation dynamics by reducing opportunities for dispersal across the landscape (Chapter 5). When there is too little movement of individuals between habitat fragments, the dwindling subpopulations within those fragments are at risk of extirpation or even extinction (Section 8.7). In contrast, well-connected subpopulations maintain themselves by colonising empty niches, exchanging genetic material, and adapting to changing environments. Dispersal also maintains critical ecosystem processes, such as pollination and seed dispersal (Section 4.2.5). Consequently, conservation biologists have invested significant resources in recent years to maintain and restore wildlife movements within fragmented ecosystems.
Connectivity in terrestrial ecosystems
Maintaining and restoring ecosystem connectivity—the ability of ecosystems to facilitate the dispersal of individuals between different areas—involves maintaining and restoring wildlife movements that are (at risk of being) impeded by human activities. The most popular method to maintain (or restore) connectivity in a fragmented landscape is to maintain (or restore) habitat linkages, also called wildlife corridors, habitat corridors, dispersal corridors, or movement corridors. All these terms refer to continuous tracts of suitable habitat with little to no dispersal barriers that connect otherwise isolated habitat patches and populations.
Maintaining and restoring ecosystem connectivity is an important strategy for conserving wildlife whose movements are impeded by human activities.
Some of the most prominent efforts to restore habitat linkages involve habitat restoration. For example, plans are currently underway to use forest regeneration to reconnect nine forest fragments in Tanzania’s East Usambara Mountain; if successful, this project would establish the largest contiguous forest block (over 3,000 km 2 ) in the Eastern Arc Mountain Biodiversity Hotspot (Newmark, 2008). The positive impact of this project is expected to be immense. It has been estimated that the restoration of just 80 km 2 of forest would stave off the first fragmentation-induced extinctions by over 2,000 years, compared to an estimated seven years until the first extinction if these forest fragments were to remain unconnected (Newmark et al., 2017).
Connectivity is important in every ecosystem on Earth. However, given the linear characteristic of riparian zones along rivers and stream—and hence a larger proportional impact of edge effects (Section 5.1.2)—we might consider connectivity in these spatially restricted systems to be particularly important (Figure 11.9). Protecting and restoring riparian zones as habitat linkages resonates with a variety of people because these areas provide a range of important ecosystem services, including flood control and water purification (Section 4.2.4). Conservationists can tap into this energy by lobbying for laws that prohibit activities such as logging, housing, and industrial developments within a certain distance from a river or stream. By protecting ecosystem services associated with riparian zones, these laws simultaneously also maintain wildlife refuges (Monadjem and Reside, 2008), source populations (Vosse et al., 2008), and habitat linkages (Bentrup et al., 2012; McLennan and Plumptre, 2012). In contrast, inadequate protection of riparian ecosystems not only compromises connectivity, but also negatively affect species not overtly dependent on these buffer areas. For example, research from Southeast Asia has shown that losing riparian ecosystems in an otherwise palm oil dominated landscape reduced stream quality, which in turn reduced local fish diversity by up to 36% (Giam et al., 2015). In contrast, protecting riparian zones were found to increase palm oil yields (Horton et al., 2018). With so many riparian areas currently being degraded and destroyed, there is an urgent need for stronger riparian protection laws (Chapter 12), and for more effective enforcement of those laws.
Restoring connectivity may also involve removing or otherwise mitigating human constructs that block wildlife dispersal. This is a major aim of TFCAs, which aim to restore dispersal between protected areas (Jones et al., 2012) by removing fences and other human constructs while still maintaining sustainable land tenures (Andersson et al., 2013). These efforts, accomplished through partnerships with local communities, are re-establishing historical mass migration routes, which in turn will hopefully also boost those areas’ ecotourism potential (Box 11.3). Efforts to revive extinct mass migrations also seem to be paying off! For example, in Botswana, the removal of veterinary fences—meant to prevent spread of diseases from wildlife to livestock, but also cutting off the world’s second largest wildebeest migration—have seen several hundred plains zebras (Equus quagga, NT) returning to old migration routes within four years (Bartlam-Brooks et al., 2011).
Simon M. Munthali
Kavango-Zambezi Transfrontier Conservation Area Programme,
Kasane, Botswana.
TFCAs are components of a larger ecosystems that straddles the border between two or more countries, encompassing one or more protected areas as well as multiple-resource areas used by communities and private landholders. They are also managed for sustainable use of natural resources (Singh, 1998). The concept recognizes that borders are political rather than ecological (Dallimer and Strange, 2015), and aims to ensure that key ecological processes continue to function where political borders have divided ecosystems, river basins, or wildlife corridors (Cumming, 1999).
TFCAs are widely being established in Africa. One of these is the 520,000 km 2 Kavango-Zambezi TFCA (KAZA)—a conservation and development initiative of Angola, Botswana, Namibia, Zambia, and Zimbabwe.
The benefits of the KAZA include:
- Re-establishment of the seasonal wildlife migration routes and connectivity among the many protected areas (national parks, community conservancies, and wildlife and forest reserves) within the region (Figure 11.C). The primary wildlife focus is the savannah elephant (Loxodonta africana), whose population of about 250,000 is predominantly concentrated in Chobe National Park (Botswana), Hwange National Park (Zimbabwe), and Bwabwata National Park (Namibia). Elephants need unimpeded movement to protected areas where population densities are much lower, such as Luengue-Luiana and Mavinga National Parks (Angola), and Sioma Ngwezi and Kafue National Parks (Zambia). This movement would reduce pressure on the ecosystems that are currently overpopulated and enable elephants and other species to better coexist—especially grazing herbivores that depend on the same habitats as the elephant.
- Expanding the wildlife-based economy, primarily ecotourism, into agricultural marginal areas (with predominantly Kalahari sand soils), through community-private partnerships. Through these partnerships, local communities would benefit from employment and business opportunities in ecotourism activities.
- Opportunities for local communities to participate in decision-making, and influencing policies and legislation related to natural management such as coordination of the fishing closed season between Namibia and Zambia during the fish breeding season (December–March) in the Zambezi River.
- Formation of alliances among different stakeholders (governments, private sector, NGOs, and local communities) to maximise skills and resources in promoting sustainable land use, conserving biodiversity and alleviating poverty.
Despite these benefits, there are obstacles to progress in attaining the benefits of the KAZA. Notable among these are social and political factors, such as increasing human population density, increasing cultivation of land, and expanding human settlements in wildlife corridors. Many of these factors trigger human-wildlife conflicts and poaching both for local consumption of bushmeat and for the illegal sale of elephant ivory. To mitigate these threats, the following strategies are being implemented:
- A Master Integrated Development Plan for the KAZA has been developed, which provides initial zoning. Its key feature is spatially allocating land into various uses (human settlement, agriculture, and protected wildlife areas, including wildlife dispersal corridors). The Master Integrated Development Plan also assists in creating awareness about the value of the wildlife corridors, which traverse communal areas.
- Promotion of conservation agriculture as a tool for improving land stewardship, intensification of agriculture, and improving crop yields per unit area of land, and therefore decreasing the likelihood of cutting down forested areas in and around wildlife corridors to plant new agricultural fields. Currently, within the KAZA, conservation agriculture is being piloted in Angola, Namibia, and Zambia. Conservation agriculture is crop production that strives to achieve acceptable profits together with high and sustained production levels while concurrently conserving the environment.
- Promotion of community-private partnerships in ecotourism development. Over the past four years, Ngoma safari lodge (Botswana), and Machenje sport fishing lodge (Zambia) have been developed specifically in support of securing wildlife corridors. They also provide incentives to the local communities for adopting wildlife conservation as a supplement to their land use practices. These lodges are in addition to the numerous existing tourist resorts in the KAZA.
- A law enforcement and anti-poaching strategy for the KAZA is being developed to coordinate transboundary law enforcement surveillance and fines to prevent poaching of protected wildlife. In addition, KAZA partner countries are integrating other security agencies, such as the military, police, immigration, and customs officials, to prevent the illegal export of wildlife products such as elephant ivory and bushmeat out of the KAZA.
- Reducing human-wildlife conflicts (Section 14.4) through improved land use planning, solar-powered electrified fencing encircling clusters of village fields and facilities and use of chilli-pepper-based olfactory repellents to deter elephants from entering crop fields.
The KAZA has made considerable progress to date in coordinating conservation efforts among the wildlife agencies and national parks across five countries in Southern Africa. The principal success has been measures to allow the continued migration of elephants along existing migration routes across international borders. The challenges ahead—from inadequate funding for wildlife patrolling and anti-poaching activities to increasing populations of rural people outside the protected areas and across migration routes—remain significant.
Section 5.1.1 discussed how inconsiderate fence placements threaten wildlife, while the paragraph above explained how removing fences can improve connectivity. Ironically, and illustrating the difficulties conservationists face when dealing with conflicting demands, strategically placed fences can sometimes also be used as a conservation tool. For example, researchers working on a fragmented lion population in Botswana found that the most effective way to improve this population’s viability was through strategic placement of fences to direct dispersal between protected areas (Cushman et al., 2016). Strategically placed predator-proof fences may at times also be required to avoid human-wildlife conflict (Packer et al., 2013, but see Creel et al., 2013), and to facilitate the recovery of threatened species, as is the case for Africa’s rarest antelope, the hirola (Beatragus hunter CR) (Ng’weno et al., 2017). The final word here is that management must remain responsive to both positive and negative impacts of tools, such as fences, rather than relegating them to bins, such as good or bad. (See also Dupuis-Desormeaux et al., [2018] for the use of fence-gaps and exclusionary fences to mitigate some negative fence impacts.)
Protecting and restoring stepping stone habitats can maintain connectivity in areas where it is impractical to establish or restore continuous habitat linkages.
At times, when it is impractical to establish or restore continuous habitat linkages, biologists may opt to protect and restore stepping stone habitats (Figure 11.10). As the name implies, stepping stone habitats are a special type of habitat linkage that facilitate dispersal along a patchwork of isolated habitat patches within a matrix of unsuitable or inhospitable habitat. Stepping stones thereby divide long dispersal events through a long stretch of inhospitable terrain up into shorter, and thus more manageable, sections. Stepping stone habitats are particularly important for migratory species that rest and refuel at stop-over sites between the end-points of their migratory route (Runge et al., 2015)—each stop-over site can be viewed as a stepping stone habitat. Prominent examples of stepping stone habitats that deserve protection include sacred forests which can act as stop-over sites for migratory forest birds; wetlands and estuaries (see Box 5.3), which can act as stop-over sites by migratory waterbirds; and small forest reserves, which can act as stepping stones between a network of other protected areas (Riggio and Caro, 2017).
Connectivity in freshwater ecosystems
Dams have always played an important role in hydropower generation and securing a year-round supply of water for farms, industries, and cities. Unfortunately, recent evidence suggests that reservoirs may create more problems than they solve (Section 5.3.2). Of concern is their contribution to greenhouse gases (Deemer et al., 2016), as well as their role in blocking dispersal of aquatic organisms. To counter these negative impacts, governments across the world are decommissioning and removing dams and other types of artificial water impoundments. For instance, over the past 30 years more than 1,174 dams were removed in the USA; the 72 dams removed in 2016 alone restored more than 3,000 km of streams (Thomas-Blate, 2016). Similar efforts are also underway in Europe ( http://www.ecrr.org ), where river restoration efforts have been initiated at over 1,100 locations across 31 countries. Unfortunately, not only are efforts to restore freshwater connectivity lagging across Africa; in many cases, even more rivers are currently being dammed (Winemiller et al., 2016).
While dams play an important role in hydropower generation and securing a year-round supply of water, recent evidence suggests that they create many environmental problems, including blocking species dispersal.
Connectivity in marine ecosystems
Ecosystem connectivity is also important in marine ecosystems. Many marine organisms, including economically important species, breed and feed in different areas at different times of the year, and use established dispersal routes to move between those areas. It is thus important to protect these dispersal routes so we can maintain these marine ecosystems and ecosystem services.
Maintaining movement dynamics in marine seascapes involves protecting and restoring marine corridors, estuarine linkages, and coastal habitat linkages.
There are three main strategies to maintain and restore movement dynamics of marine seascapes. First, marine corridors—zones used by whales and other marine species to move between feeding and breeding grounds—should be protected. Marine biologists in several countries successfully reduced collisions between whales and ocean-faring vessels with minor adjustments to shipping lanes that previously crossed marine corridors (Silber et al., 2012). Second, estuarine linkages should be protected, and restored where needed. For example, biologists in South Africa restored the natural flow regime of the St Lucia Estuary, Africa’s largest estuarine lake, by removing dredge spoil in the estuary mouth (Nunes et al., 2018). Third, coastal habitat linkages—beaches and littoral shallows used by wildlife for dispersal, breeding, and feeding—need to be maintained. Studies from South Africa have highlighted how poor protection of connectivity pathways between coastal habitats can compromise these areas’ high levels of species richness and endemism (von der Heyden, 2009; Harris et al., 2014).
Mimicking connectivity
In the absence of habitat linkages, wildlife managers may be able to mimic dispersal dynamics by sporadically translocating a few individuals between subpopulations. Managing populations in this way may be a good alternative in cases where areas earmarked for translocations are too small to sustain a single viable population. Such is the case in South Africa, where conservation biologists occasionally move threatened predators within a small and fragmented protected areas network, where none of the areas are large enough to host a viable population on their own (see Box 8.3). Managing isolated and small populations so intensively nearly always requires sound underlying principles and extensive quantitative analyses (Chapter 9) for guidance.
Management considerations in connectivity conservation
While intuitively appealing, there are a few potential drawbacks to connectivity that conservation planners should consider when planning to establish new habitat linkages (reviewed in Haddad et al., 2014). Prominently, connecting historically isolated populations may lead to outbreeding depression, for example when populations with different local adaptations are connected. Habitat linkages may also act as bottlenecks that expose dispersing animals to greater risks of predation and enable pests and diseases to spread easier. Care must be taken to ensure that wildlife do indeed perceive the landscape “connected”; a habitat linkage that may look good to the human eye may in fact be perceived as inhospitable habitat to wildlife (Newmark, 2008). A recent study from the Americas has shown that the habitat quality of a single stepping stone habitat can determine whether a migration is successful or not (Gómez et al., 2017).
Although the benefits for reconnecting fragmented landscapes generally outweigh the drawbacks, it is important to carefully plan to avoid those drawbacks.
Although the benefits for connecting landscapes for conservation generally outweigh the drawbacks (Haddad et al., 2014), it is important to carefully plan to avoid those drawbacks. Genetic studies can be useful in both determining connectivity among populations (von der Heyden, 2009; Godley et al., 2010) and help researchers detecting potential deleterious factors, such as outbreeding depression (Figure 11.11, see also Frankham et al., 2011; Ralls et al., 2018). Modelling approaches that combine a target species’ movement limitations with radio tracking technologies (e.g. Godley et al., 2010) or remotely sensed environmental variables (e.g. Wegmann et al., 2014) could help to estimate whether a landscape is indeed connected. Much effort has also been invested in finding the optimal width of habitat corridors. For example, one study in lowland forests suggested that corridors that are 30–40 m wide might be adequate for migration of most species while corridors that are 200 m wide will be adequate for all species (Laurance and Laurance, 1999). This is useful guidance, but ecosystems vary, as do target species (Wilson et al., 2010; Pryke and Samways, 2012) and, thus, some corridors may need to be even wider.


14.6: Social Media and Retail
- Explain why social media is a crucial part of any current retailer’s IMC mix
Social Media can be a powerful element of the IMC Mix for three (3) important reasons:
- Consumers opt-in to engage with the firm, its brand or products
- The firm, its brand or products can optimize their message for the channel
- Social Media allows for two-way communication and a multiplier effect
The importance of the first factor cannot be over-stated. Consumers who opt-in to engage with a brand communicate a level of interest that marketers cannot determine through traditional media. That is, when a consumer “comes to you” and expresses interest, marketers should be confident that they have an interested prospect who can be developed into a customer.
Second, social media can provide platforms for the firm, its brand or products to communicate in a natural way with consumers and its customers. That is, firm owns the medium, whether Twitter, Instagram or Facebook. Thus, they can curate their content for more nuanced or expansive messages. Unlike television or radio or even print, which have physical limitations, e.g. time or space, social networks allow firm and brands to message in the way that best suits them, e.g. long form storytelling, video, imagery. Further, brands can express their personality, confident that the majority of their social media followers are loyal customers and supporters. Think about the positive reaction Wendy’s has gotten from their Twitter activity.
The final consideration is that social media supports the consumer, giving them a platform with which to engage the firm or brand directly. In fact, one of the emerging benefits of Twitter is that firms are using it as a Customer Service platform, monitoring critiques, complaints issues in real-time to resolve concerns. But, this doesn’t have to be limited to current customers and failures of service. Instead, potential customers can also use social media to ask questions and solicit feedback from current customers. This is where the multiplier effect comes into play. There is the organized firm/ brand response, but also the content creation from loyal and engaged consumers.
Contributors and Attributions
- Social Media and Retail. Authored by : Patrick Williams. Provided by : Lumen Learning. License : CC BY: Attribution