Answer
Solution set = $\{12\}$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{lllll}
x-6 \gt 0, & and & x-4 \gt 0 & and & x \gt 0\\
x \gt 6 & & x \gt 4 & &
\end{array}\right]$
$ x \gt 6 \qquad (*)$
- is the condition eventual solutions must satisfy.
$\log_{2}(x-6)+\log_{2}(x-4)-\log_{2}x=2$
... LHS: apply the product rule, ... RHS write $2$ as $\log_{2}2^{2}$
$\log_{6}[(x-6)(x-4)]-\log_{2}x=\log_{2}2^{2}$
... LHS: apply the quotient rule,
$\displaystyle \log_{6}\frac{(x-6)(x-4)}{x}=\log_{2}2^{2}$
... if $\log_{b}M=\log_{b}N,$ then M=N
$\displaystyle \frac{(x-6)(x-4)}{x}=4\quad $ ... multiply with x
$ x^{2}-10x+24=4x $
$ x^{2}-14x+24=0$
$(x-12)(x-2)=0$
$ x=2$ - does not satisfy (*), discard
$ x=12$ - satisfies (*)
Thus, the solution set = $\{12\}$
