<s>[INST] Solving $\frac{3^x+2^x}{3^x-2^x}=7$: More than one answer? How to solve for all $x$? 
I am trying to solve this problem (math for fun):
$$\frac{3^x+2^x}{3^x-2^x}=7$$

Step 1. Let$\:a=3^x\:and\:b=2^x$
Step 2. $\frac{a+b}{a-b}=7$
Step 3. $a+b=7\left(a-b\right)=7a-7b$
Step 4. $6a-8b=0$
Step 5. $6a=8b=3\cdot 2\cdot a=4\cdot 2\cdot b$
Step 6. $3a=4b$
Step 7. $3\cdot 3^x=2\cdot 2\cdot 2^x=3^{x+1}=2^{x+2}$
Step 8. Log both sides
Step 9. $\ln \left(3\right)+x\ln 3=x\ln2+\ln 4$
Step 10. $\ln \left(3\right)-\ln 4=x\:\left(\ln 2-\ln 3\right)$
Step 11. $\frac{\:\left(\ln 3-\ln 4\right)}{\left(\ln 2-\ln 3\right)\:}=x≈0.71$
SOLVED^
 [/INST] To answer you question how do you know how many solutions there are and how to you find them all:  By keeping track of your steps and assuring you never add extraneous solution and that you steps are all one to one reversable so you don't lose or add solutions.
So $\frac {3^x + 2^x}{3^x -2^x} = 7$ gives us the condition that $3^x \ne 2^x$.
Substitute $a =3^x$ and $b=2^x$.  This is just substitution. But we temporarily lose all information we may have about $3^x, 2^x$.  But we don't have to worry we will get the information back.
$\frac {a+b}{a-b} = 7\implies a+b = 7(a-b)$. This adds an extraneous solution where $a = b$ which we know can not happens.  (This is actually some of the information we lost when substitute.  We know $2^x, 3^x > 0$ so $2^x + 3^x > 0$ so $a-b =0$ is not possible... okay lets not worry.  We'll have to keep track $a \ne b$.
$8b = 6a$ just adding to each side.  Completely reversible. This doesn't affect our solutions.
$4b = 3a$ ditto.  This is a restriction.
$4\cdot 2^x = 3\cdot 3^x$
$2^{x+2} = 3^{x+1}$.  We know longer are dealing with $a,b$ so the caviet $a\ne b$ no longer is a concern.
$\ln 2^{x+2} = \ln 3^{x+1}$.  As $\ln$ is one to one this keeps the number of solutions the same.
$(x+2)\ln 2 = (x+1)\ln 3$  Linear shifting doesn't affect anything so
$x(\ln 2 - \ln 3) = (\ln 3 -2\ln 2)=(\ln 3-\ln 4)$.  As $\ln 2-\ln 3$ is a constant and $\ln 2 \ne\ln 3$ we can divide both sides and that doesn't affect the number of solutions.
And $x -\frac {\ln 3 - \ln 4}{\ln 2-\ln 3}$.
Nothing we did lost any solutions and the one thing that added extraneous solution took care of itself.
So we know this is the only solution.
BTW $\frac {\ln 3-ln 4}{\ln 3- \ln 2} = \frac {\ln \frac 34}{\ln \frac 32}$ or $\frac {\ln 3 -2\ln 2}{\ln 3 - \ln 2} = 1 -\frac {\ln 2}{\ln 3 - \ln 2}= 1-\frac {\ln 2}{\ln \frac 32}$
If is helps (probably doesn't) $\frac {\ln \frac 34}{\ln \frac 32} =\log_{\frac 32} \frac 34=\log_{\frac 32} \frac 32\cdot \frac 12 = 1 - \log_{\frac 32} 2$
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