How do you evaluate # log 25 + 1/2log3#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Adrian D. Aug 6, 2015 #log(25*sqrt(3))=1.64# 2 d.p. Explanation: Using: #loga+logb=logab# and #bloga=loga^b# => #log25+1/2log3=log(25*sqrt(3))=1.64# 2 d.p. 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 2560 views around the world You can reuse this answer Creative Commons License