Question #1e6f1

Jun 21, 2015

The solution is : $x = e \vee x = \frac{1}{e}$

Explanation:

The logarythms have different bases, so first you have to change base using formula:

${\log}_{a} x = {\log}_{b} \frac{a}{\log} _ b x$

In this case we get:

${\log}_{e} x - \frac{1}{\log} _ e x = 0$

${\left({\log}_{e} x\right)}^{2} - 1 = 0$

$\left({\log}_{e} x - 1\right) \left({\log}_{e} x + 1\right) = 0$

${\log}_{e} x - 1 = 0 \vee {\log}_{e} x + 1 = 0$

${\log}_{e} x = 1 \vee {\log}_{e} x = - 1$

$x = e \vee x = \frac{1}{e}$