Question

The value of $x$ satisfying ${\log }_2(3x-2)={\log }_{\frac{1}{2}}x$ is equal to

• A. $-\frac{1}{3}$
• B. $2$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

Answer: (D)

$1$

${\log }_2(3x-2)={\log }_{\frac{1}{2}}x$
Above equation is valid, when $x>\frac{2}{3}$
${\log }_2(3x-2)={\log }_{\frac{1}{2}}x$
$\Rightarrow {\log }_2(3x-2)=-{\log }_{2}x,[\because {\log }_{\frac{1}{a}}b=-{\log }_{a}b]$
$\Rightarrow {\log }_2(3x-2)={\log }_2\frac{1}{x},[\because x\log a=\log a^x]$
$\Rightarrow 3x-2=\frac{1}{x}$
$\Rightarrow 3x^2-2x-1=0$
$\Rightarrow x=1,-\frac{1}{3}$
Since, $x>\frac{2}{3}$
$\therefore x=1$
Ans: D