#sum _(a,b,c) (1/ (1+ log _a bc)) = 1 #Prove it? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Ananda Dasgupta Mar 21, 2018 Since #log_a b = log b/log a#, we have #1/(1+log_a bc) = 1/(1+(log (bc))/log a) = log a/(log a+log (bc)) = log a/log(abc)# Thus, the sum is #log a/log(abc) + log b/log(abc) + log c/log(abc) = (log a +log b+ log c)/log(abc) = log(abc)/log(abc) = 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 3200 views around the world You can reuse this answer Creative Commons License