How do you solve #sqrt7^(6x)= 49^(x-6)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Apr 14, 2016 #x = -12# Explanation: Replace #sqrt7# by #7^(1/2)# and 49 by #7^2.#. Now, use #(a^m)^n=a^(mn)#. #7^(3x)=7^(2x-12)#, Compare exponents.. #3x=2x-12#. #x=-12#. 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 1401 views around the world You can reuse this answer Creative Commons License