# Solve lnx = 1-ln(x+2) for x?

## Thanks in advance

May 4, 2018

$x = \sqrt{1 + e} - 1 \approx 0.928$

#### Explanation:

Add $\ln \left(x + 2\right)$ to both sides to get:
$\ln x + \ln \left(x + 2\right) = 1$

Using the addition rule of logs we get:
$\ln \left(x \left(x + 2\right)\right) = 1$

Then by $e \text{^}$ each term we get:
$x \left(x + 2\right) = e$

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

$x = \frac{- 2 \pm \sqrt{{2}^{2} + 4 e}}{2}$

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

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

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

$x = - 1 \pm \sqrt{1 + e}$

However, with the $\ln \left(\right)$s, we can only have positive values, so $\sqrt{1 + e} - 1$ can be taken.

May 4, 2018

$x = \sqrt{e + 1} - 1$

#### Explanation:

lnx=1−ln(x+2)
$A s 1 = \ln e$
$\implies \ln x = \ln e - \ln \left(x + 2\right)$

$\ln x = \ln \left(\frac{e}{x + 2}\right)$
Taking the antilog on both sides,
$x = \frac{e}{x + 2}$
$\implies {x}^{2} + 2 x = e$
Complete the squares.
$\implies {\left(x + 1\right)}^{2} = e + 1$
$\implies x + 1 = \pm \sqrt{e + 1}$
$\implies x = \sqrt{e + 1} - 1 \mathmr{and} x = - \sqrt{e + 1} - 1$

We neglect the second value as it would be negative, and the logarithm of a negative number is undefined.