How do you solve #log_2x=log_5 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria May 26, 2016 #x=1.605# Explanation: #log_2x=log_53# can be simplified using #log_ba=loga/logb#. Hence it is #logx/log2=log3/log5# or #logx=log3/log5xxlog2# or #logx=0.4771/0.6990xx0.3010# Hence #x=10^(0.4771/0.6990xx0.3010)=1.605# 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 1188 views around the world You can reuse this answer Creative Commons License