How do you solve #log_3 x = 9 log_x 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis Nov 18, 2015 Using the change of base rule you get #logx/log3=9log3/logx=>(logx)^2=(3log3)^2=> (logx-3log3)*(logx+3log3)=0=> x=3^3=27 or x=1/27# 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 6997 views around the world You can reuse this answer Creative Commons License