How do you solve #log 64 = 2logx#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jun 25, 2018 #x=8# Explanation: #log64# = #log2^6# = #6log2# = #2(3log2)# = #2log2^3# = #2log8# But we are given #log64=2logx# Hence #x=8# 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 4302 views around the world You can reuse this answer Creative Commons License