How do you solve for x in #32^x=6#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub May 4, 2016 #x=0.517# Explanation: #log 32^x = log 6# #x log 32 = log 6# #x = log 6/log 32# #x=0.517# 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 972 views around the world You can reuse this answer Creative Commons License