How do you solve #(1/4)^x=64#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer MG May 3, 2018 #x=-3# Explanation: #(1/4)^x=64# #1/4^x=64# #1=64*4^x# #1=4^3*4^x# #1/4^3=4^x# #4^-3=4^x# #-3=x# 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 6564 views around the world You can reuse this answer Creative Commons License