# How do you solve -3e^(9x-1)+6=-58?

Nov 20, 2016

$x = \frac{1}{9} \left(\ln \left(\frac{64}{3}\right) + 1\right)$

#### Explanation:

$- 3 {e}^{9 x - 1} + 6 = - 58$

Subtract 6 from each side:
$- 3 {e}^{9 x - 1} = - 64$

Divide each side by -3:
${e}^{9 x - 1} = \frac{64}{3}$

To isolate x, rewrite the equation as a natural log:
$9 x - 1 = \ln \left(\frac{64}{3}\right)$

$9 x = \ln \left(\frac{64}{3}\right) + 1$
$x = \frac{1}{9} \left(\ln \left(\frac{64}{3}\right) + 1\right)$
$x = \frac{1}{9} \left(\ln 64 - \ln 3 + 1\right)$