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How do you solve $ln x + ln (x + 1) = ln 12$ $lnx + ln(x+1) = ln12$ ?

1 Answer

Oct 23, 2015

$x = 3$ $x=3$

Explanation:

In general
$XXX log (A) + log (B) = log (A B)$ $color(white)("XXX")log(A)+log(B) = log(AB)$
(this is one of the basic logarithmic rules)

Specifically
$XXX ln (x) + ln (x + 1) = ln (x^{2} + x)$ $color(white)("XXX")ln(x)+ln(x+1) = ln(x^2+x)$
and we are told, this is
$XXXXXXXXXXXXX = ln (12)$ $color(white)("XXXXXXXXXXXXX")=ln(12)$

$\Rightarrow x^{2} + x = 12$ $rArr x^2+x = 12$

$XXX x^{2} + x - 12 = 0$ $color(white)("XXX")x^2+x-12=0$

$XXX (x + 4) (x - 3) = 0$ $color(white)("XXX")(x+4)(x-3)=0$

$XXX x = - 4$ $color(white)("XXX")x=-4$ $XXX$ $color(white)("XXX")$ or $XXX x = 3$ $color(white)("XXX")x=3$

Since $ln (x)$ $ln(x)$ is not defined for negative values of $x$ $x$
$\Rightarrow x = 3$ $rArr x=3$

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