# How do you solve 2 ln x+ ln x^2=3?

Dec 4, 2015

$x = {e}^{\frac{3}{4}}$

#### Explanation:

First of all, you need to "unite" the $\ln$ expressions into one.

This can be done with the logarithmic rules:

${\log}_{a} \left(n\right) + {\log}_{a} \left(m\right) = {\log}_{a} \left(n \cdot m\right)$

$r \cdot {\log}_{a} \left(n\right) = {\log}_{a} \left({n}^{r}\right)$

So, you can transform your equation as follows:

$\textcolor{w h i t e}{\times} 2 \ln x + \ln {x}^{2} = 3$

$\iff 2 \ln x + 2 \ln x = 3$

$\iff 4 \ln x = 3$

... divide both sides by $4$ ...

$\iff \ln x = \frac{3}{4}$

Now, the inverse function for $\ln x$ is ${e}^{x}$ which means that both $\ln \left({e}^{x}\right) = x$ and ${e}^{\ln x} = x$ hold.

This means that you can apply ${e}^{x}$ to both sides of the equation to "get rid" of the logarithm:

$\iff {e}^{\ln} \left(x\right) = {e}^{\frac{3}{4}}$

$\iff x = {e}^{\frac{3}{4}}$