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?

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.
