How do you solve #-log x = 4.176#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Dec 7, 2015 #x~= 6.668xx10^(-5)# Explanation: #-log x = 4.176# #color(white)("XXX") rArr log x = -4.176# #color(white)("XXX")10^(-4.176) = x# therefore (using a calculator) #color(white)("XXX")x = 6.668xx10^(-5)# Remember #color(white)("XXX")log_b a = c# means #b^c = a# and #color(white)("XXX")#the default base (#b#) for #log# is #10# 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 1300 views around the world You can reuse this answer Creative Commons License