How do you evaluate #log_x(1/36)= -(2/3)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Mar 10, 2016 #x=216# Explanation: #x^(-2/3) =1/36# #x^(-2/3) =6^-2# #(x^(-2/3))^(-3/2) =(6^-2)^(-3/2)# #x=6^3=216# 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 3528 views around the world You can reuse this answer Creative Commons License