Answer
The solution set is $\{12\}$.
Work Step by Step
The given equation is
$\log_2(x-6)+\log_2(x-4)-\log_2x=2$
Use product rule.
$\log_2(x-6)(x-4)-\log_2x=2$
Use quotient rule.
$\log_2 \frac{(x-6)(x-4)}{x}=2$
Write the equation in exponent form.
$ \frac{(x-6)(x-4)}{x}=2^2$
$ \frac{(x-6)(x-4)}{x}=4$
Multiply both sides by $x$
$x\cdot \frac{(x-6)(x-4)}{x}=x\cdot 4$
Simplify.
$(x-6)(x-4)= 4x$
Use the distributive property on the left hand side.
$ x^2 -6x-4x+24= 4x$
Subtract $4x$ from both sides.
$ x^2 -6x-4x+24-4x= 4x-4x$
Simplify.
$ x^2 -14x+24= 0$
Factor.
$ x^2 -12x-2x+24= 0$
$ x(x -12)-2(x-12)= 0$
$ (x -12)(x-2)= 0$
Set each factor equal to zero.
$ x -12=0 $ or $x-2= 0$
Isolate $x$.
$ x =12 $ or $x=2$
Check the solution.
For $x=12$
$\log_2(12-6)+\log_2(12-4)-\log_212=2$
$\log_2(6)+\log_2(8)-\log_212=2$
$\log_2(6\cdot8)-\log_212=2$
$\log_2\left (\frac{6\cdot8}{12} \right )=2$
$\log_2\left (4 \right )=2$
$\log_2\left (2^2 \right )=2$
$2=2$ True.
For $x=2$
$\log_2(2-6)+\log_2(2-4)-\log_22=2$
$\log_2(-4)+\log_2(-2)-\log_22=2$
Not a solution.