How do you evaluate #g(x)=-lnx# for #x=1/2#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer Binayaka C. May 16, 2017 For # x=1/2; g(x) ~~0.693147# ;# Explanation: #x= 1/2 , g(x) = - ln(x) :. g(1/2) = -ln(1/2) = -[ln(1)-ln(2)] ~~ - [0-0.693147) ~~0.693147# For # x=1/2; g(x) ~~0.693147# ;# [Ans] 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 1342 views around the world You can reuse this answer Creative Commons License