Answer
The solution set is $\{3\}$
Work Step by Step
The given equation is
$\log_2(x-1)-\log_2(x+3)=\log_2\left ( \frac{1}{x} \right )$
Use quotient rule.
$\log_2 \left ( \frac{x-1}{x+3}\right )=\log_2\left ( \frac{1}{x} \right )$
$\frac{x-1}{x+3}= \frac{1}{x} $
Multiply both sides by $x(x+3)$.
$x(x+3)\cdot \frac{x-1}{x+3}= x(x+3)\cdot \frac{1}{x} $
Simplify.
$x(x-1)= (x+3)$
$x^2-x= x+3$
Subtract $x+3$ from both sides.
$x^2-x-(x+3)= x+3-(x+3)$
$x^2-x-x-3= x+3-x-3$
$x^2-2x-3= 0$
Factor.
$x^2+x-3x-3= 0$
$x(x+1)-3(x+1)= 0$
$(x+1)(x-3)= 0$
Set both factors equal to zero.
$x+1=0$ or $x-3 =0$
Isolate $x$.
$x=-1$ or $x=3$
Check for the solution.
For $x=-1$.
$\log_2(-1-1)-\log_2(-1+3)=\log_2\left ( \frac{1}{-1} \right )$
$\log_2(-2)-\log_2(2)=\log_2\left ( -1 \right )$
Not a solution.
For $x=3$.
$\log_2(3-1)-\log_2(3+3)=\log_2\left ( \frac{1}{3} \right )$
$\log_2(2)-\log_2(6)=\log_2\left ( \frac{1}{3} \right )$
$\log_2\left (\frac{2}{6}\right)=\log_2\left ( \frac{1}{3} \right )$
$\log_2\left (\frac{1}{3}\right)=\log_2\left ( \frac{1}{3} \right )$