How do you expand #Log_b( sqrt57/74)#? Precalculus Properties of Logarithmic Functions Functions with Base b 1 Answer A. S. Adikesavan May 20, 2016 #((1/2) log 57 - log 74)/log b=((1/2) ln 57 - ln 74)/ln b# Explanation: Use #log_b a=log_c a/log_c b and log_b a^n=n log_b a#. Here, the given logarithm #= log_b( sqrt57 /74)# #=log( sqrt57 /74)/log b=ln( sqrt57 /74)/ln b# #=( log 57/2 - log 74)/log b=( ln 57/2 - ln 74)/ln b# Answer link Related questions What is the exponential form of #log_b 35=3#? What is the product rule of logarithms? What is the quotient rule of logarithms? What is the exponent rule of logarithms? What is #log_b 1#? What are some identity rules for logarithms? What is #log_b b^x#? What is the reciprocal of #log_b a#? What does a logarithmic function look like? How do I graph logarithmic functions on a TI-84? See all questions in Functions with Base b Impact of this question 1436 views around the world You can reuse this answer Creative Commons License