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How do you solve $log (5 x - 6) = 2 log x$ $log(5x-6) = 2log x$ ?

1 Answer

May 21, 2016

$x = 2$ $x=2$ or $x = 3$ $x=3$

Explanation:

Note that
$XXX 2 log (x) = log (x^{2})$ $color(white)("XXX") 2log(x)=log(x^2)$

So
$XXX log (5 x - 6) = 2 log (x)$ $color(white)("XXX")log(5x-6)=2log(x)$
is equivalent to
$XXX log (5 x - 6) = log (x^{2})$ $color(white)("XXX")log(5x-6)=log(x^2)$

which implies
$XXX 5 x - 6 = x^{2}$ $color(white)("XXX")5x-6=x^2$

$XXX x^{2} - 5 x + 6 = 0$ $color(white)("XXX")x^2-5x+6=0$

$XXX (x - 2) (x - 3) = 0$ $color(white)("XXX")(x-2)(x-3)=0$

So either
$XXX (x - 2) = 0 XX \to XX x = 2$ $color(white)("XXX")(x-2)=0color(white)("XX")rarrcolor(white)("XX")x=2$
or
$XXX (x - 3) = 0 xx \to XX x = 3$ $color(white)("XXX")(x-3)=0color(white)("xx")rarrcolor(white)("XX")x=3$

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