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

Oct 30, 2017

$\frac{3 + \sqrt{13}}{2}$

#### Explanation:

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

If:

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

Then:

$\frac{x + 1}{x - 2} = x = {x}^{2} - 3 x - 1$

Solving: ${x}^{2} - 3 x - 1$

$x = \frac{3 + \sqrt{13}}{2} \mathmr{and} x = \frac{3 - \sqrt{13}}{2}$

$\frac{3 - \sqrt{13}}{2}$ not valid as gives negative value in $\ln \left(x - 2\right)$

So solution is:

$\frac{3 + \sqrt{13}}{2}$

Oct 30, 2017

$x = \frac{3}{2} + \frac{\sqrt{13}}{2}$

#### Explanation:

Taking the exponent of both sides, we get:

$\frac{x + 1}{x - 2} = x$

Multiplying both sides by $\left(x - 2\right)$ we get:

$x + 1 = {x}^{2} - 2 x$

Subtract $x + 1$ from both sides to get:

$0 = {x}^{2} - 3 x - 1$

$\textcolor{w h i t e}{0} = {\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4} - 1$

$\textcolor{w h i t e}{0} = {\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{13}}{2}\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(x - \frac{3}{2} - \frac{\sqrt{13}}{2}\right) \left(x - \frac{3}{2} + \frac{\sqrt{13}}{2}\right)$

So:

$x = \frac{3}{2} \pm \frac{\sqrt{13}}{2}$

One of these values $\frac{3}{2} + \frac{\sqrt{13}}{2}$ is positive and a valid solution of the original equation.

The other value $\frac{3}{2} - \frac{\sqrt{13}}{2}$ is negative so not a solution if $\ln$ is the real logarithm. In fact it is not even a solution if we consider the complex logarithm, for which:

$\ln \left(\frac{3}{2} - \frac{\sqrt{13}}{2} + 1\right) - \ln \left(\frac{3}{2} - \frac{\sqrt{13}}{2} - 2\right)$

$= \ln \left(\frac{5}{2} - \frac{\sqrt{13}}{2}\right) - \ln \left(\frac{1}{2} + \frac{\sqrt{13}}{2}\right) - i \pi$

$\ne \ln \left(\sqrt{\frac{13}{2}} - \frac{3}{2}\right) + i \pi = \ln \left(\frac{3}{2} - \frac{\sqrt{13}}{2}\right)$

Oct 30, 2017

See below.

#### Explanation:

In the quest for real solutions, supposing $x \ne \left\{0 , 2\right\}$

${\log}_{e} \left\mid x + 1 \right\mid - {\log}_{e} \left\mid x - 2 \right\mid = {\log}_{e} \left\mid x \right\mid$ or

${\log}_{e} \left(\frac{\left\mid x + 1 \right\mid}{\left\mid x - 2 \right\mid \left\mid x \right\mid}\right) = 0$ or

$\frac{\left\mid x + 1 \right\mid}{\left\mid x - 2 \right\mid \left\mid x \right\mid} = 1$ whose solutions are included in the solutions for

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

$\left(x + 1 - \left(x - 2\right) x\right) \left(x + 1 + \left(x - 2\right) x\right) = 0$ or

$\left(1 + 3 x - {x}^{2}\right) \left(1 - x + {x}^{2}\right) = 0$

and the real solutions are from

$1 + 3 x - {x}^{2} = 0$ with the only feasible solution

$x = \frac{1}{2} \left(3 + \sqrt{13}\right)$