How do you solve #log x^2 + log 25 = 2#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis Sep 11, 2015 The answer is #x=2# Explanation: We have that #logx^2+log25=2=>log25x^2=log100=>25x^2=100=>x^2=4=>x=2,x=-2# But #x>0# hence #x=2# 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 4104 views around the world You can reuse this answer Creative Commons License