How do you solve #log_2 x=log_5 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer George C. Aug 16, 2015 #x = 2^(ln(3)/ln(5)) ~~ 1.605# Explanation: By the change of base formula #log_5(3) = ln(3)/ln(5)# Then #x = 2^(log_2 x) = 2^(log_5(3)) = 2^(ln(3)/ln(5))# 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 1413 views around the world You can reuse this answer Creative Commons License