How do you solve #Log_a X = 3 Log_a 2 #? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer SaltĀ² Jun 22, 2018 x=8 Explanation: just apply #a*log(b)=log(b^a)# then cancel the #log#s and you get #x=2^3# or #x=8# 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 1441 views around the world You can reuse this answer Creative Commons License