How do you solve #log_5 x + log_5 6 = 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer mason m Nov 27, 2015 #x=25/6# Explanation: Know that: #log_ab+log_ac=log_abc#. #log_5x+log_5 6=2# #log_5 6x=2# #5^2=6x# #x=25/6# 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 1350 views around the world You can reuse this answer Creative Commons License