How do you solve 2^x=5? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k May 6, 2016 x=log5/log2~~2.32 Explanation: Given' 2^x=5 Taking log of both sides =>log2^x=log5 =>xlog2=log5 =>x=log5/log2=0.6989/0.3010~~2.32 Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve 9^(x-4)=81? How do you solve logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 2? How do you solve 2 log x^4 = 16? How do you solve 2+log_3(2x+5)-log_3x=4? See all questions in Logarithmic Models Impact of this question 26844 views around the world You can reuse this answer Creative Commons License