How do you evaluate #log_(1/13) (1/169)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Roy E. Jan 20, 2017 #2# Explanation: Spotting that #169=13^2# so you need #log_(1/13)(1/13^2)#. Since #log_a(x^2)=2log_a(x)# and #log_a(a)=1# you get #log_(1/13)((1/13)^2)=2log_(1/13)(1/13)=2 xx 1 = 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 1772 views around the world You can reuse this answer Creative Commons License