How do you solve #5-log _ 6 (2x - 6) + log _ 6 x = 2#?

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Answer 1 · 2018-03-19T13:18:01

#5−log_6(2x−6)+log_6x=2#

#5-2=log_6(2x−6)-log_6x#

#3=log_6((2x−6)/x)# [Since #color(red)(loga-logb = log(a/b)#]

#6^3 = ((2x−6)/x)#

#(2x−6)=216x#

#−6=214x#

#x=-3/107#