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!
