How do you evaluate #log_(2/5) (125/8)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Vinay K. · Stefan V. Nov 24, 2016 #- 3# Explanation: Useful Hint for this question: #log_aa^n=nlog_aa=n# Solution : #log_(2/5)(125/8)=log_(2/5)(5/2)^3# = #log_(2/5)(2/5)^(-3)# =#(-3)log_(2/5)(2/5)# = -3. [As #log _a a=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 4346 views around the world You can reuse this answer Creative Commons License