How do you solve for x in $log_3(x^(2)-2x)=1$ ?

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Answer 1 · 2015-12-31T13:42:21

$x_1=-1$ and $x_2=3$

$log_3 (x^2-2x)=1$
$x^2-2x=3^1$
$x^2-2x-3=0$
$Delta = 4 +12 =16$
$x=(-b+-sqrt(Delta))/2=(2+-4)/2$ => $x_1=-1$ and $x_2=3$

How do you solve for x in log_3(x^(2)-2x)=1log_3(x^(2)-2x)=1?