How do you solve #(1/4)^(2x)=(1/32)^(3x+1)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Cesareo R. May 26, 2016 #x = -5/11# Explanation: #(1/4)^{2x}=(1/32)^{3x+1} equiv 1/(4)^{2x}=1/(32)^{3x+1}# # 1/(4)^{2x}=1/(32)^{3x+1} equiv (4)^{2x}=(32)^{3x+1} # # (4)^{2x}=(32)^{3x+1} equiv (2^2)^{2x}=(2^5)^{3x+1}# #(2^2)^{2x}=(2^5)^{3x+1}equiv 2^{2 times 2 x}=2^{5 times(3x+1)}# # 2^{2 times 2 x}=2^{5 times(3x+1)}equiv 4x=15x+5# solving for #x# we get #x = -5/11# 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 1709 views around the world You can reuse this answer Creative Commons License