How do you solve #(5^x)^(x+1)=5^(2x+12)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Mar 9, 2016 #x=-3# or #x=4# Explanation: As #(a^m)^n=a^(mn)#, #(5^x)^(x+1)=5^(x(x+1))=5^(x^2+x)# Hence, #5^(x^2+x)=5^(2x+12)# or #x^2+x=2x+12# or #x^2+x-2x-12=0# or #x^2-x-12=0# or #x^2-4x+3x-12=0# or #x(x-4)+3(x-4)=0# or #(x+3)(x-4)=0# or #x=-3# or #x=4# 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 1655 views around the world You can reuse this answer Creative Commons License