<s>[INST] Solve the equation $\log_2(\frac{8}{2^x}-1)=x-2$ This equation comes from the book of Prilepko. It looks sufficiently simple but I want to know if my solution is correct.
$\log_2(\frac{8}{2^x}-1)=x-2$
I have only been able to transform into this:
$\log_2(\frac{8-2^x}{2^x})=x-2$
$\iff$ $\log_2(8-2^x)=2x-2$
$\iff$ $8-2^x=2^{2x-2}$
$\iff$ $32-4.2^x-2^{2x}= 0$
Let $t=2^x$
$-t^2-4t+32=0$
This equation will produce $t=-8$ and $t=4$
Therefore $x=2$.
Another question that I wish to ask is when solving these logarithmic equations, one very often substitute a dummy variable to overcome the restriction caused by the order of operations. For example, you cannot do much directly with $5^{2x}-130.5^x+625$. But let $t=5^x$, then you can easily factorize the this equation into $(t-125)(t-5)=0$, hence $(5^x-125)(5^x-5)=0$. Isn't this a sort of manipulating symbols according to certain prescribed rules? Is there any logical difficulty with this kind of operation? Since by substituting a new variable, it means that this variable must inherit all properties which the object of the substitution originally possesses. For example, if $a^x$ is the object of substitution for $t$, then if $a<0$ and x can assumed a form of rational power, doesn't this make no sense at all?
Do you know any example in elementary mathematics where this method of substitution can yield contradictory result?
 [/INST] If you want "to stretch" the logarithm method or solution to the very end here it is:
$$\log_{2} \left ( \frac{8}{2^{x}} - 1 \right ) = x-2 $$
$$\log_{2} \left ( \frac{8-2^{x}}{2^{x}} \right ) = x-2 $$
$$\log_{2} \left (8-2^{x} \right ) - \log_{2} \left ( 2^{x} \right ) = x-2 $$
$$\log_{2} \left (2^{3}-2^{x} \right ) - x = x-2 $$
$$\log_{2} \left (2^{3}-2^{x} \right ) = 2x-2 $$
$$\log_{2} \left (2^{3} \left ( 1- 2^{x-3}\right ) \right ) = 2x-2 $$
$$\log_{2} \left (2^{3}\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$3\log_{2} \left (2\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$3 + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2 $$
$$ \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-5 $$
Using the $\textrm{Antilogarithm}$ (see here or here for any details)
$$ \textrm{antilog}_{2} \left ( \log_{2} \left ( 1- 2^{x-3} \right ) \right ) = \textrm{antilog}_{2} \left ( 2x-5 \right ) $$
$$1- 2^{x-3}= 2^{2x-5}$$
Rearranging the above equation translates into:
$$1- 2^{x}\times 2^{-3}= 2^{2x}\times 2^{-5}$$
$$2^{2x}\times 2^{-5} - 2^{x}\times 2^{-3} - 1 = 0$$
Multiplying by $2^{5}$ becomes into the expression you had found:
$$2^{2x} - 2^{x}\times 2^{2} - 32 = 0$$
From then on it is consistent with a quadratic equation:
$$A=2^{x}$$
$$A_{1,2}=\frac{-4\pm \sqrt{16+128}}{2}=\frac{-4\pm \sqrt{144}}{2}$$
$$A_{1,2}=\frac{-4\pm 12}{2}\,,A_{1}=\frac{-4+12}{2}=4\,,A_{2}=\frac{-4-12}{2}=-8$$
Since we cannot obtain negative results from a power the second solution is discarded. Therefore:
$$2^{x}=4$$
$$x=2$$
Becoming the result you had just found. So it is correct. As mentioned this is more or less the same of what others before me had just commented.
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