How do you solve #log_5 2x − 5 = −4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Bdub May 4, 2016 #x=1563/625# Explanation: Use definition #log_b x=y iff b^y=x# #5^-4 = 2x-5# #1/5^4 = 2x-5# #1/625 = 2x-5# #1/625 + 5 = 2x# #3126/625 = 2x# #x=1563/625# 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 1963 views around the world You can reuse this answer Creative Commons License