How do you solve #8(4^(6-2x))+13=41#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jul 19, 2017 #x~=2.548# Explanation: #8(4^(6-2x))+13=41# #=>8(4^(6-2x))=41-13=28# or #4^(6-2x)=28/8=7/2# or #2^(12-4x)=7/2=3.5# Therefore #12-4x=log_2 3.5=log3.5/log2# or #12-4x=0.54407/0.30103=1.807355# i.e. #4x=12-1.807355=10.192645# and #x=10.192645/4~=2.548# 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 2410 views around the world You can reuse this answer Creative Commons License