How do you solve the equation #log_7 24-log_7(y+5)=log_7 8#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria Dec 2, 2016 #y=-2# Explanation: #log_7 24-log_7 (y+5)=log_7 8# #hArrlog_7(24/(y+5))=log_7 8# hence #24/(y+5)=8# or #24=8xx(y+5)# or #y+5=24/8=3# and #y=3-5=-2# 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 1670 views around the world You can reuse this answer Creative Commons License