How do you solve #2^x = 5^(x - 2) #? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Mar 4, 2016 #x=3.513# Explanation: To solve #2^x=5^(x−2)#, take logs of both the sides, this becomes #xlog2=(x-2)log5=xlog5-2log5# and transposing terms this becomes #xlog5-xlog2=2log5# or #x(log5-log2)=2log5# #x=(2log5)/(log5-log2)=(2xx0.699)/(0.699-0.301)# or #x=1.398/0.398=3.513# 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 1768 views around the world You can reuse this answer Creative Commons License