Logx.log2x=log4x solve equatin?

May 5, 2018

$x = 2$

Explanation:

As per the question, we have :

$\log x . \log 2 x$ = $\log 4 x$

$\log x . 2 x = \log 4 x$

$\log 2 {x}^{2} = \log 4 x$

$2 {x}^{2} = 4 x$

$2 {x}^{2} - 4 x = 0$

${x}^{2} - 2 x = 0$

$\left(x\right) \left(x - 2\right) = 0$

$\therefore$ $x = 0 \mathmr{and} 2$

But $x$ cannot be $0$ as $\log 0$ is not defined.

$\therefore$ $x = 2$