Question

The value of x satisfying $ log_2(3x-2) = log{_{1/2}} x $ is

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

Answer 1

Answer: (A)

Given, $log_2(3x - 2) = log_{\frac{1}{2}} x$
$\Rightarrow \frac{log(3x - 2)}{log\,2} = \frac{log\,x}{log\,\frac{1}{2}} = \frac{log\,x}{-log\,2}$
$\Rightarrow log(3x - 2) = log\,\frac{1}{x}$
$\Rightarrow 3x^2 - 2x - 1 = 0$
$\Rightarrow (x - 1)(x + \frac{1}{3}) = 0$
$ \Rightarrow x = 1, x = - \frac{1}{3}$
But $ x = - \frac{1}{3}$ is not satisfied in $log_{\frac{1}{2}} x$.
Hence, $ x = 1 $ is the required solution.