Question

If ${\log }_{4}8+{\log }_4(x+3)-{\log }_4(x-1)=2,$ then $x=$

Answer 1

$x+3>0,x-1>0\Rightarrow x>1$

${\log }_{4}8+{\log }_4(x+3)-{\log }_4(x-1)=2$
$\Rightarrow {\log }_4\frac{8(x+3)}{x-1}=2$
$\Rightarrow \frac{8(x+3)}{x-1}=4^2$
$\Rightarrow x+3=2x-2$
$\Rightarrow x=5$