How do you solve #8^(-2-x)=431#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Feb 5, 2017 #x=-4.9172# Explanation: As from definition of log #a^m=b# implies #log_a b=m# Also #log_a b=logb/loga# where #log# without mentioning base means logarithm to the base #10# and we can get this from tables. #8^(-2-x)=431# means #log_8 431=-2-x# or #-2-x=log431/log8=2.6345/0.9031=2.9172# Hence, #x=-2-2.9172=-4.9172# 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 2058 views around the world You can reuse this answer Creative Commons License