How do you solve # (¼)log_6 (a – 3) – log_6 3 = 0#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer maganbhai P. Jul 26, 2018 #a=84# Explanation: We have , #1/4log_6(a-3)-log_6 3=0# #=>1/4log_6(a-3)=log_6 3# #=>log_6(a-3)=4log_6 3to["Multiplying both sides by 4]"# #=>log_6(a-3)=log_6 3^4to[becauselog_k x^n=nlog_k x]# #=>log_6(a-3)=log_6 81# #=>a-3=81# #=>a=81+3# #=>a=84# 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 1132 views around the world You can reuse this answer Creative Commons License