How do you solve $log_6(4x+4)=log_6 64$ ?

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Answer 1 · 2016-11-22T07:54:44

$x=15$

$log_6 (4x+4)=log_6 64$

Hence $log_6 (4x+4)-log_6 64=0$

or $log_6 ((4x+4)/64)=0=log_6 1$

or $(4x+4)/64=1$ i.e. $4x+4=64$

and $x=(64-4)/4=60/4=15$

How do you solve log_6(4x+4)=log_6 64log_6(4x+4)=log_6 64?