How do you solve #25^x-28=97#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Jul 30, 2016 #x=1.5# Explanation: #25^x-28=97# or #25^x=97+28# or #25^x=125# or #(5^2)^x=5^3# or #5^(2x)=5^3# or #2x=3# and #x=3/2=1.5# 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 932 views around the world You can reuse this answer Creative Commons License