How do you solve $(\ln ( x ) ) ^ { 2} + 2\ln ( x ) = + 15$ ?

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Answer 1 · 2016-11-17T16:00:26

$x=e^((-5))color(white)(aaa)x=e^3$
or
$x=.0067379color(white)(aaa)x=20.0855$

$(ln(x))^2 +2ln(x)=15$

Subtract 15 from both sides.

$(lnx)^2+2lnx-15=0$

Factor (just factor as if the equation were $x^2+2x-15=0$ , then use $lnx$ in place of $x$ ).

$(lnx+5)(lnx-3)=0$

Set each factor equal to zero and solve.

$lnx+5=0color(white)(aaa)lnx-3=0$

$lnx=-5color(white)(aaa)lnx=3$

Rewrite as an exponential. The base of $ln$ is $e$ .

$x=e^((-5))color(white)(aaa)x=e^3$

$x=.0067379color(white)(aaa)x=20.0855$

How do you solve (\ln ( x ) ) ^ { 2} + 2\ln ( x ) = + 15(\ln ( x ) ) ^ { 2} + 2\ln ( x ) = + 15?