How do you solve Log 5 = 2 - Log x? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer José F. Feb 27, 2016 x=e^2/5 Explanation: log5=2−logx logx=2−log5 logx= log (e^2)-log5 logx= log (e^2/5) x=e^2/5 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 2274 views around the world You can reuse this answer Creative Commons License