How do you solve #5^x=17#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer ali ergin Mar 24, 2016 #x=1,7603744273# Explanation: #5^x=17# #log 5^x=log 17# #x*log 5=log 17# #log 5=0,6989700045# #log 17=1,2304489214# #x=(log 17)/(log 5)# #x=1,7603744273# 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 9153 views around the world You can reuse this answer Creative Commons License