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

##### 1 Answer

May 31, 2018

Take the natural logarithm of both sides:

#ln(e^(3c- 2)) = lnb#

#(3c- 2)ln e = lnb#

#3c - 2 = lnb#

#c= 1/3(lnb + 2)#

Hopefully this helps!