# How do you solve Ln(x+1)-ln(x-2)=lnx^2?

Apr 10, 2018

#### Answer:

approximately:
$x = 2.5468$

#### Explanation:

${\ln}^{\frac{x + 1}{x - 2}}$ = ${\ln}^{{x}^{2}}$
we can cancel out the (Ln) parts and the exponents would be left out;
$\frac{x + 1}{x - 2}$ = ${x}^{2}$
$x + 1$ = ${x}^{2.} \left(x - 2\right)$
$x + 1$ = ${x}^{3} - 2 {x}^{2}$
${x}^{3}$-$2 {x}^{2}$-$x$-$1$=$0$
$x = 2.5468$