How do you solve #25^x = 5^(x^2 - 15)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer ali ergin Mar 28, 2016 #x={-3,5}# Explanation: #25^x=5^(x^2-15)# #color(red)(5)^(2x)=color(red)(5)^(x^2-15)# #"equation has the same base value"# #2x=x^2-15# #x^2-2x-15=0# #(x-5)(x+3)=0# #if (x-5)=0" then " x=5# #if(x+3)=0" then "x=-3# #x={-3,5}# 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 1101 views around the world You can reuse this answer Creative Commons License