How do you solve $logx+log(x+21)=2$ ?

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Answer 1 · 2016-11-23T06:14:46

$x=25$

$logx(x+21)=2$
$log(x^2+21x)=2$
$x^2+21x=10^2$
$x^2+21x-100=0$
$(x-4)(x+25)=0$
$x=4$ and $x=-25$
$x=-25$ is extraneous
$x=4$

How do you solve logx+log(x+21)=2logx+log(x+21)=2?