How do you solve #4^ { 3x - 9} - 10= - 3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Carl S. Mar 28, 2018 #x = {log_4(7) + 9}/3 approx 4.4# Explanation: #4^{3x -9} - 10 = -3# #4^{3x -9} = 7# //add 10 to both sides #3x -9 = log_(4) 7# //#log_4# on both sides (#log_a(a^{x})=x#) #3x = log_4(7) + 9# //add 9 to both sides #x = {log_4(7) + 9}/3# //divide both sides by 3 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 1270 views around the world You can reuse this answer Creative Commons License