How do you solve #(1/2)^(3x)=1/16#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Noah G Jun 28, 2016 An alternative method... Explanation: #(1/2)^(3x)= (1/2)^4# #2^(-3x) = 2^-4# #-3x = -4# #x = 4/3# Hopefully this helps! 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 1430 views around the world You can reuse this answer Creative Commons License