How do you solve #log_6 (x-4) ^2=2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Guillaume L. May 9, 2018 #x_"1"=4+6^sqrt2# #x_"2"=4+1/(6^sqrt2)# Explanation: #log_"6"(x-4)²=2# #(ln(x-4)/ln(6))²=2# #ln(x-4)/ln(6)=±sqrt2# #ln(x-4)=±ln(6)sqrt2# #x-4=6^(±sqrt2)# #x=4+6^(±sqrt2)# #x_"1"=4+6^sqrt2# #x_"2"=4+1/(6^sqrt2)# \0/ here's our answer! 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 1390 views around the world You can reuse this answer Creative Commons License