Question

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

Answer 1

${\log }_{4}8+{\log }_4\left(x+3\right)-{\log }_4\left(x-1\right)=2$

$={\log }_4\frac{8\left(x+3\right)}{x-1}=2$

$=\frac{8\left(x+3\right)}{x-1}=4^2$

$=\frac{8\left(x+3\right)}{x-1}=16$

$x+3=2x-2$

$x=5$