How do you calculate #Log_sqrt(17)152#? Precalculus Properties of Logarithmic Functions Functions with Base b 1 Answer A. S. Adikesavan Aug 3, 2016 #2 (log 152/log 17)= 3.5464#, nearly. Explanation: Use #log_b a=log a/log b#. #log_sqrt 17 152# #=log 157/logsqrt 17# #=log 167/log(17^(1/2)) #=2 (log 152/log 17)=3.5464, #nearly. 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 1404 views around the world You can reuse this answer Creative Commons License