# How do you solve lnx+ln(2x)=2?

Aug 6, 2016

$x = \frac{\sqrt{2}}{2} e$

#### Explanation:

If $x > 0$ then:

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

and

$2 = \ln {e}^{2}$

So we have:

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

Since $\ln$ is one-one as a Real valued function of positive Reals, this implies that:

$2 {x}^{2} = {e}^{2}$

So:

${x}^{2} = {e}^{2} / 2$

Hence:

$x = \frac{e}{\sqrt{2}} = \frac{\sqrt{2}}{2} e$

Note we ignore the negative square root since we are only looking at the case $x > 0$.