How do you solve #4logx= log (1/16)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Mar 17, 2016 #x=1/2# Explanation: As #4logx=log(1/16)# #logx^4=log(1/2^4)# or #x^4=2^(-1xx4)=(2^(-1))^4# Hence #x=2^(-1)# or #x=1/2# 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 1825 views around the world You can reuse this answer Creative Commons License