Question

Find the value of $x$ which satisfies the expression ${\log }_2(3x-2)={\log }_{\frac{1}{2}}x$

Answer 1

${\log }_2(3x-2)={\log }_{\frac{1}{2}}x=\frac{{\log }_{2}x}{{\log }_2{2}^{-1}}={\log }_2{x}^{-1}[\because {\log }_{b}a=\frac{\log a}{\log b}\text{\&}m\log a=\log a^m\text{\&}{\log }_{a}a=1]$
$\Rightarrow$ $3x-2=x^{-1}$
$\Rightarrow$ $3x^2-2x=1$
$\Rightarrow$ $x=1$ or $x=-\frac{1}{3}$.
But ${\log }_2(3x-2)$ and ${\log }_{\frac{1}{2}}x$ are meaningful if $x>\frac{2}{3}$.
Hence, $x=1$
Ans: 1