# How do you solve e^ { 3c - 2} = b?

May 31, 2018

Take the natural logarithm of both sides:

$\ln \left({e}^{3 c - 2}\right) = \ln b$

$\left(3 c - 2\right) \ln e = \ln b$

$3 c - 2 = \ln b$

$c = \frac{1}{3} \left(\ln b + 2\right)$

Hopefully this helps!