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

Dec 14, 2015

$x = \sqrt{\frac{{e}^{3}}{6}}$

#### Explanation:

Use the fact that $\ln \left(a b\right) = \ln \left(a\right) + \ln \left(b\right)$ for positive $a$ and $b$.

$\ln 3 x + \ln 2 x = \left(\ln 3 + \ln x\right) + \left(\ln 2 + \ln x\right)$

$= \ln 6 + 2 \ln x$

$= 3$

Perform algebraic manipulation yields

$\ln x = \frac{3 - \ln 6}{2}$

$x = {e}^{\frac{3 - \ln 6}{2}}$

$= \frac{{e}^{\frac{3}{2}}}{{e}^{\frac{\ln 6}{2}}}$

$= \sqrt{\frac{{e}^{3}}{6}}$