How do you evaluate #log_(1/11) (1/121)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Somebody N. Apr 5, 2018 #color(blue)(2)# Explanation: Using change of base formula: #log_(1/11)(1/121)=(ln(1/121))/ln(1/11)# #log(a/b)=loga-logb# #(ln(1/121))/ln(1/11)=(ln(1)-ln(121))/(ln(1)-ln(11))=(0-ln(121))/(0-ln(11))=ln(121)/ln(11)# #loga^b=bloga# #=(ln(11^2))/ln(11)=(2ln(11))/ln(11)=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 2599 views around the world You can reuse this answer Creative Commons License