How do you solve for x in #(125)^x = 625#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Apr 9, 2016 #x=4/3=1.333# Explanation: #125^x=625# means #x=log_(125)625# As #log_ba=loga/logb# #x=log625/log125=2.79588/2.09691=1.333# Alternately - #125^x=625hArr(5^3)^x=5^4hArr5^(3x)=5^4# or #3xlog5=4log5# or #x=4/3=1.333# 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 3557 views around the world You can reuse this answer Creative Commons License