How do you solve #log_4 (x-6)^3= 6#?

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Answer 1 · 2015-11-07T21:29:29

#x=22#

By using laws of logs we can write this equation as :

#3log_4(x-6)=6#

Now divide throughout by 3 to get

#log_4(x-6)=2#

Now applying the definition of log, we get

#x-6=4^2#

#therefore x=22#