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How do you solve ${log}_{2} (x^{2} - 10) = {log}_{2} (3 x)$ $log_2(x^2 - 10)= log_2 (3x)$ ?

1 Answer

Apr 8, 2016

$x = 5$ $x=5$

Explanation:

${log}_{b} p = {log}_{b} q$ $log_b p = log_b q$
$\Rightarrow p = q$ $rArr p=q$

Therefore
$XXX {log}_{2} (x^{2} - 10) = {log}_{2} (3 x)$ $color(white)("XXX")log_2(x^2-10)=log_2(3x)$
$XXX \Rightarrow x^{2} - 10 = 3 x$ $color(white)("XXX")rArr x^2-10=3x$

$XXX \to x^{2} - 3 x - 10 = 0$ $color(white)("XXX")rarr x^2-3x-10=0$

$XXX \to (x - 5) (x + 2) = 0$ $color(white)("XXX")rarr (x-5)(x+2)=0$

$XXX \to x = 5 XX or XX x = - 2$ $color(white)("XXX")rarr x=5color(white)("XX")orcolor(white)("XX")x=-2$

But neither $log$ $log$ is defined when $x = - 2$ $x=-2$ and therefore this result is extraneous.

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