How do you solve #11^(x^2) =25.4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Mar 17, 2016 #x=1.1614# or #x=-1.1614# Explanation: Taking log of both the sides in the equation #11^(x^2)=25.4#, #x^2xxlog11=log25.4# or #x^2=(log25.4)/log11=1.4048/1.0414=1.349# Hence #x=+-1.1614# i.e. #x=1.1614# or #x=-1.1614# 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 1295 views around the world You can reuse this answer Creative Commons License