How do you solve log_2 x=log_5 3log2x=log53? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer George C. Aug 16, 2015 x = 2^(ln(3)/ln(5)) ~~ 1.605x=2ln(3)ln(5)≈1.605 Explanation: By the change of base formula log_5(3) = ln(3)/ln(5)log5(3)=ln(3)ln(5) Then x = 2^(log_2 x) = 2^(log_5(3)) = 2^(ln(3)/ln(5))x=2log2x=2log5(3)=2ln(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)=819x−4=81? How do you solve logx+log(x+15)=2logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=2? How do you solve 2 log x^4 = 162logx4=16? How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=4? See all questions in Logarithmic Models Impact of this question 1464 views around the world You can reuse this answer Creative Commons License