How do you solve $log_3(x^2) - log_3(x+3) =3$ ?

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Answer 1 · 2015-12-20T16:23:16

$x= (27+-9sqrt5)/2$

LHS= $log_3 ((x^2)/(x+3))$ , Hence it is $log_3 ((x^2)/(x+3))$ =3

$((x^2)/(x+3))= 3^3 =27$

$x^2 -27x -81=0$

$x= (27+-sqrt (729-324))/2$

$x= (27+-9sqrt5)/2$

How do you solve log_3(x^2) - log_3(x+3) =3 log_3(x^2) - log_3(x+3) =3?