How do you simplify #Log_(2)(10)#? Precalculus Properties of Logarithmic Functions Logarithm-- Inverse of an Exponential Function 1 Answer George C. Aug 30, 2015 #log_2(10) = 1/log(2)# Explanation: Use the change of base formula: #log_a(b) = log_c(b)/log_a(b)# #log_2(10) = log_10(10)/log_10(2) = 1/log(2)# since #log_10 = log# In general, #log_a(b) = log_b(b)/log_b(a) = 1/log_b(a)# 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 2406 views around the world You can reuse this answer Creative Commons License