How do you solve #log y = log (x-1) + 1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer A. S. Adikesavan Mar 25, 2016 y = #10 (x-1)#, x > 1. Explanation: Working on common logarithm, #y = 10^((log(x-1)+1)#. #10^(m+n)= 10^m 10^n# and #10^(log a)=a#. #y = (x-1) X 10^1=10(x-1)# Here, #log(x-1)# is defined for x > 1. So, the answer is suject yo x > 1.. 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 3359 views around the world You can reuse this answer Creative Commons License