How do you solve #5^(x-3) = 21#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Apr 21, 2016 #x=ln21/ln5 +3~~4.89# Explanation: #ln 5^(x-3)=ln21# Use Property #log_bx^n=nlog_bx# #(x-3) ln5=ln21# #x-3=ln21/ln5# #x=ln21/ln5 +3~~4.89# 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 1076 views around the world You can reuse this answer Creative Commons License