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

Sep 22, 2016

$x = 1.729$

#### Explanation:

$\ln \left(x - 1\right) + \ln \left(x + 2\right) = 1$

$\Leftrightarrow \ln \left(x - 1\right) \left(x - 2\right) = 1$ or

$\left(x - 1\right) \left(x + 2\right) = e$ or

${x}^{2} + x - 2 - e = 0$ and using quadratic fomula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \times 1 \times \left(- 2 - e\right)}}{2}$

= $\frac{- 1 \pm \sqrt{1 + 8 + 4 e}}{2}$

= $\frac{- 1 \pm \sqrt{9 + 4 e}}{2}$

= $\frac{- 1 \pm \sqrt{9 + 4 \times 2.7183}}{2}$

= $\frac{- 1 \pm \sqrt{9 + 10.8732}}{2}$

= $\frac{- 1 \pm \sqrt{19.8732}}{2}$

= $\frac{- 1 \pm 4.4579}{2}$

i.e. $x = 1.729$ or $x = - 2.729$

But $x$ cannot have a negative value, hence $x = - 2.729$