How do you solve #1.2^x -1= 104#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Harish Chandra Rajpoot Jul 4, 2018 #x=25.526# Explanation: #1.2^x-1=104# #1.2^x=104+1# #1.2^x=105# Taking log on both the sides #\log(1.2^x)=\log(105)# #x\log(1.2)=\log(105)# #x=\frac{\log(105)}{\log(1.2)}# #x=25.526# 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 1405 views around the world You can reuse this answer Creative Commons License