How do you solve #\frac { 1} { 5} = 5^ { x - 1}#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer LM Dec 18, 2016 #x = 0# Explanation: #1/a^m = a^-m# #1/5 = 1/5^1# #=5^-1# substitute power for #x - 1#: #x - 1 = -1# add 1: #x = 0# 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 857 views around the world You can reuse this answer Creative Commons License