If #5^(x+2)=4#, find #x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Mar 14, 2017 #x=-1.1386# Explanation: As #5^((x+2))=4# #(x+2)log5=log4# i.e. #x+2=log4/log5=0.60206/0.69897# i.e. #x+2=0.8614# and hence #x=0.8614-2=-1.1386# 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 1083 views around the world You can reuse this answer Creative Commons License