How do you solve #log x^2+1=5#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub Mar 21, 2016 #x=100# Explanation: #log x^2 +1 =5# #logx^2=4#-> subtract 1 from both sides #2logx=4#->Use property #log_bx^n=n*log_b x# #log x=2#-> divide both sides by 2 #10^2=x#->use property #y=log_bx iff b^y=x# #x=100# 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 2819 views around the world You can reuse this answer Creative Commons License