How do you solve #16^(x-5) = 9^(x+1)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Apr 21, 2016 #x=(ln1048576+ln9)/(ln16-ln9)~~27.91# Explanation: #ln 16^(x-5)=ln9^(x+1)# #(x-5)ln16=(x+1)ln9# #xln16-5ln16=xln9+ln9# #xln16-ln16^5=xln9+ln9# #xln16-ln 1048576=xln9+ln9# #xln16-xln9=ln1048576+ln9# #x(ln16-ln9)=ln1048576+ln9# #x=(ln1048576+ln9)/(ln16-ln9)~~27.91# 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 1421 views around the world You can reuse this answer Creative Commons License