How do you solve #log_6(4x+4)=log_6 64#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Shwetank Mauria Nov 22, 2016 #x=15# Explanation: #log_6 (4x+4)=log_6 64# Hence #log_6 (4x+4)-log_6 64=0# or #log_6 ((4x+4)/64)=0=log_6 1# or #(4x+4)/64=1# i.e. #4x+4=64# and #x=(64-4)/4=60/4=15# Answer link Related questions What is a logarithm? What are common mistakes students make with logarithms? How can a logarithmic equation be solved by graphing? How can I calculate a logarithm without a calculator? How can logarithms be used to solve exponential equations? How do logarithmic functions work? What is the logarithm of a negative number? What is the logarithm of zero? How do I find the logarithm #log_(1/4) 1/64#? How do I find the logarithm #log_(2/3)(8/27)#? See all questions in Logarithm-- Inverse of an Exponential Function Impact of this question 2301 views around the world You can reuse this answer Creative Commons License