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

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

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Answer 1 · 2016-11-23T06:14:46

$x = 25$ $x=25$

Explanation:

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

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

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