How do you solve #-4=log_5x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Tony B Dec 8, 2015 #x=1/625# Explanation: Consider #log_10 x=y -> 10^y=x# Using the same concept for your question: #log_5x=(-4) -> 5^(-4)=x# so # x= 1/(5^4) =1/625# 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 1200 views around the world You can reuse this answer Creative Commons License