How do you solve #4^(2-5x)=1/64#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Konstantinos Michailidis May 29, 2016 Notice #1/64=4^-3# hence #4^(2-5x)=4^-3=>log(4^(2-5x))=log4^-3=>(2-5x)=-3=>5=5x=>x=1# 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 1566 views around the world You can reuse this answer Creative Commons License