## Precalculus (6th Edition) Blitzer

$x=4$
$\log_{2} x+\log_{2} (x-2)=0 \implies 3=\log_{2} 2^3$ Re-arrange as: $x(x-2)=2^3$ $\implies x^2-2x-8=0$ $\implies (x^2+2x)-(4x+8)=0$ or, $(x+2)(x-4)=0$ So, $x=-2,4$ Neglect the negative term of $x$, because it does not belong to the domain of the variable; so we have $x=4$