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

Jul 21, 2018

$\textcolor{b l u e}{x = \frac{- 1 + \sqrt{4e+9}}{2} \approx 1.7290}$

#### Explanation:

Using the logarithmic property:

$\ln a + \ln b = \ln a b$

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

Using the antilogarithm:

${e}^{\ln \left(x - 1\right) \left(x + 2\right)} = {e}^{1}$

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

Expanding $L H S$

${x}^{2} + x - 2 - e = 0$

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

$x = \frac{- 1 + \sqrt{4e+9}}{2} \approx 1.7290$

$x = \frac{- 1 - \sqrt{4e+9}}{2} \approx - 2.7290$

Checking solutions with original equation:

$\ln \left(1.7290 - 1\right) + \ln \left(1.7290 + 2\right) = 1.000058554$

$\ln \left(- 2.7290 - 1\right) + \ln \left(- 2.7290 + 2\right) \setminus \setminus \textcolor{red}{X}$

This is undefined for real numbers.

$\ln x$ is only defined for real numbers if $x > 0$

$x = \setminus \frac{- 1 + \setminus \sqrt{9 + 4 e}}{2} = 1.72896$

#### Explanation:

$\setminus \ln \left(x - 1\right) + \setminus \ln \left(x + 2\right) = 1 \setminus \quad \left(\setminus \forall \setminus \setminus x > 1\right)$

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

$\setminus \ln \left({x}^{2} + x - 2\right) = \setminus \ln e$

Comparing the numbers on same base on both the sides,

${x}^{2} + x - 2 = e$

${x}^{2} + x - \left(2 + e\right) = 0$

Solving above quadratic equation as follows

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

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

But, $x > 1$ hence

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

$= 1.72896$