How do you solve #400(1-0.2)^x=50#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Guillaume L. May 9, 2018 #x=ln(8)/ln(5/4)# Explanation: #400(1-0.2)^x=50# #ln(400(1-0.2)^x)=ln50# #lnab=lna+lnb# #ln400+ln(0.8^x)=ln50# #lna^b=blna# #ln400+xln0.8=ln50# #xln0.8=ln(50/400)# #xln0.8=ln(1/8)# #cancel(-)xln(5/4)=cancel(-)ln8# #x=ln(8)/ln(5/4)# \0/ here's our answer! 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 1265 views around the world You can reuse this answer Creative Commons License