Question

Find the value of $x$ satisfying the equation $\frac{1}{{\log }_6(x+3)}+\frac{2{\log }_{0.25}(4-x)}{{\log }_{0.5}(x+3)}=0$.

Answer 1

Answer: 3

${\log }_{x+3}6-{\log }_{x+3}(x+3)+\frac{2{\log }_{0.5}(4-x)}{\left({\log }_{0.5}0.25\right){\log }_{0.5}(x+3)}=0$
${\log }_{x+3}\left(\frac{6}{x+3}\right)+\frac{{\log }_{0.5}(4-x)}{{\log }_{0.5}(x+3)}=0$
$\Rightarrow {\log }_{x+3}\left(\frac{6}{x+3}\right)+{\log }_{x+3}(4-x)=0$
$\Rightarrow {\log }_{x+3}\left(\frac{6(4-x)}{x+3}\right)=0\Rightarrow \frac{6(4-x)}{x+3}=1$
$\Rightarrow 24-6x=x+3\Rightarrow 7x=21\Rightarrow x=3$