How to solve $\log_2(x)+\log_{10}(x-7)=3$ using high-school math? A question given in a grade 12 "advanced functions" class, asks to solve $\log_2(x)+\log_{10}(x-7)=3$ with a hint to change bases.
The given hint suggests the base of the second logarithm is 10, but when trying to massage the equation, how does one proceed after reaching $x(x-7)^{\log(2)}=8$ using only high school methods?
It is possible to "cheat" by observing that $\log(1) = 0$ for any base, and that $\log_2(8) = 3$, so $x=8$ is a solution, but that only works because of the fortunate selection of constants.
What is missing?
 A: The only thing you need to prove is that $x=8$ is the only solution.
Proof. $x\le 7$ is impossible because then the second $\log$ is not defined. If $7<x<8$ then the first $\log$ is $<3$ and the second $\log$ is $<0$, so the sum is $<3$. Finally if $x>8$ then the first $\log$ is $>3$ and the second $\log$ is $>0$, so the sum is $>3$. So the only possibility left is $x=8$. $\Box$
A: The exact solution can be made.
My goal here is to completely eliminate logarithmic expressions and reveal the "algebraic meat" of the equation.

$$\log_2x+\log_{10}(x-7)-\log_28-\log_{10}1=0 \Longrightarrow \log_2{\left(\frac x8\right)}=\log_{10}\left(\frac{1}{x-7}\right) \Longrightarrow \dfrac{\ln {\left(\frac x8\right)}}{\ln 2}=\dfrac{\ln\left(\frac{1}{x-7}\right) }{\ln 10} \Longrightarrow \dfrac {\ln {\left(\frac x8\right)}}{\ln\left(\frac{1}{x-7}\right) }=\log_2{10}$$
where, $x≠8.$

Then we have:

$$\begin{cases} \ln {\left(\frac x8\right)}=k\ln 10 \\ \ln\left(\frac{1}{x-7}\right)=k \ln 2 \\ k≠0 \end{cases} \Longrightarrow \begin{cases} \dfrac x8=e^{k\ln 10} \\ \dfrac{1}{x-7}=e^{k\ln2} \\ k≠0 \end{cases} \Longrightarrow 8×e^{k\ln 10}=\dfrac{1}{e^{k\ln2}}+7, \thinspace k≠0$$


Finally, we get the transcendental equation!
$$\color{red}{\boxed{2^{k + 3} ×5^k= 2^{-k} + 7}}$$ where $k≠0.$

As far as I know, there is no closed form for our last equation and there is no solution in the real number set.
$k=0$ is our only real solution. It is possible to see clearly that it cannot be $k> 0$ and $k <0.$
Remark.
However, our last equation is actually equivalent to our original equation.  Because the last transcendental equation also works when $ k = 0 $, which implies $x=8.$
