Question

Find the value of $-4x$, if $x$ satisfies ${\log }_{(1-x)}\left(3\right)-{\log }_{(1-x)}\left(2\right)=\frac{1}{2}$

Answer 1

${\log }_{1-x}a$ is valid when $1-x>0,1-x\ne 1$
$\Rightarrow x<1,x\ne 0$
${\log }_{1-x}\left(3\right)-{\log }_{1-x}\left(2\right)=\frac{1}{2}$
$\Rightarrow {\log }_{1-x}\left(\frac{3}{2}\right)[\because \log a-\log b=\log \frac{a}{b}]=\frac{1}{2}\Rightarrow \frac{3}{2}={(1-x)}^{\frac{1}{2}}$
$\Rightarrow \frac{9}{4}=1-x\Rightarrow x=1-\frac{9}{4}=\frac{-5}{4}$
Therefore, $-4x=5$
Ans: 5