Question

Solve the following equation.
$2{\log }_4(4-x)=4-{\log }_2(-2-x)$

Answer 1

$2{log}_4(4-x)=4-{log}_2(-2-x)$

Domain of given equation is $4-x>0$ and $-2-x>0$

$x<4$ and $x<-2$

Therefore, $x\in (-\infty ,-2)$

${log}_{2}4-x=4-{log}_2[-(x+2\left)\right]\Rightarrow 4-x=\frac{16}{-(x+2)}\Rightarrow (x-4)(x+2)=16\Rightarrow x^2-2x-24=0\Rightarrow x=6,-4$

As $x=6$ is not in domain

Therefore, $x=-4$ is the solution.