How do you solve #9^(2y) = 66#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Mar 28, 2016 #y=(1/2)(log 66/log 9)# =0.9534, nearly. Explanation: Equating logarithms, #log(9^(2y))=log 66#. So, 2y log 66 = log 9, using #log(a^m)=mlog a#.. #y=(1/2)(log 66/log 9)# 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 1965 views around the world You can reuse this answer Creative Commons License