# Question #158af

Oct 21, 2017

$x = 1$

#### Explanation:

$\ln \left(a\right) + \ln \left(b\right) = \ln \left(a b\right)$

So:

$\ln \left({x}^{3}\right) + \ln \left(\frac{2}{x}\right) = \ln \left(\frac{2 {x}^{3}}{x}\right) = \ln \left(2 {x}^{2}\right)$

If:

$\ln \left(2 {x}^{2}\right) = \ln \left(2\right)$

Then:

$2 {x}^{2} = 2 \implies x = \pm \sqrt{1}$

Only $x = 1$ is valid, since logarithms of negative numbers are undefined.

As a note:

You could have solved this just using the clue you were given.

$2 \ln x = 0$

$\ln x = 0$

This means that if e is raised to the power of 0 it equals x.

${e}^{0} = x$

${e}^{0} = 1$

So:

$x = 1$