How do you solve for x in $log_6x=2-log_6 4$ ?

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Answer 1 · 2018-07-03T14:09:44

$color(blue)(x=9)$

$log_6(x)=2-log_6(4)$

$log_6(x)+log_6(4)=2$

$log(a)+log(b)=log(ab)$

$log_6(4x)=2$

Raising the base to these:

$6^(log_6(4x))=6^2$

$4x=6^2$

$x=(6^2)/4=36/4=9$

How do you solve for x in log_6x=2-log_6 4log_6x=2-log_6 4?