# How do you solve ln(-3x-1)-ln7=2?

Jul 25, 2016

$x = - \frac{7 {e}^{2} + 1}{3}$

#### Explanation:

Since $\log a - \log b = \log \frac{a}{b}$,

you can rewrite the equation as:

$\ln \left(\frac{- 3 x - 1}{7}\right) = 2$

you can note that $2 = \ln {e}^{2}$, then

$\ln \left(\frac{- 3 x - 1}{7}\right) = \ln {e}^{2}$

or

$\frac{- 3 x - 1}{7} = {e}^{2} \mathmr{and} - 3 x - 1 > 0$

$- 3 x - 1 = 7 {e}^{2} \mathmr{and} - 3 x > 1$

$- 3 x = 7 {e}^{2} + 1 \mathmr{and} 3 x < - 1$

$x = - \frac{7 {e}^{2} + 1}{3} \mathmr{and} x < - \frac{1}{3}$

that's true