How do you write #log_7 (1/49)=-2# in exponential form? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer salamat Jan 27, 2017 see explaination Explanation: Let say #log_7(1/49)=y# #(1/49) = 7^y# use concept if #log_ab=c#, then #b=a^c# #1/7^2=7^y# #7^-2=7^y# #y=-2# therefore, #log_7(1/49)=-2# 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 3595 views around the world You can reuse this answer Creative Commons License