How do you solve #2log x = 4log 7#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Dec 24, 2015 #x=49# Explanation: Remember: #color(white)("XXX")a*log(c) = log(c^a)# Therefore #color(white)("XXX")2log(x)=4log(7)# #rarrcolor(white)("XXX")log(x^2)=log(7^4)# #rarrcolor(white)("XXX")x^2=7^4# #rarrcolor(white)("XXX")x=7^2=49# 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 1300 views around the world You can reuse this answer Creative Commons License