How do you solve #log_3 75 = 2x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k Mar 27, 2016 #log_3 75=2x=>3^(2x)=75=>9^x=75=>xlog9=log75# #=>x=log75/(log3^2)=log75/(2log3)# 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 1596 views around the world You can reuse this answer Creative Commons License