How do you solve #e^(3x)=20#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k May 14, 2016 #~~1# Explanation: #e^(3x)=20# Taking #log_e# on both sides we have #e^(3x)=20# #=>log_ee^(3x)=log_e20# #=>(3x)log_ee=log_e20 " "# using formula #log_em^n=nlog_em# #=>(3x)=log_e20 " "# since #log_ee=1# #=>x=log_e20/3 ~~3/3=1# Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 4064 views around the world You can reuse this answer Creative Commons License