# How do you solve lnx + ln(x+4) = 6?

Oct 29, 2015

$- 2 \pm \sqrt{4 + {e}^{6}}$

#### Explanation:

Start of by rewriting the left hand side using properties of logarithms

$\ln \left(x \left(x + 4\right)\right) = 6$

Distribute the $x$

$\ln \left({x}^{2} + 4 x\right) = 6$

Rewrite both sides in terms of the base $e$

${e}^{\ln \left({x}^{2} + 4 x\right)} = {e}^{6}$

Rewrite left hand side using properties of logarithms

${x}^{2} + 4 x = {e}^{6}$

Substract ${e}^{6}$ form both sides

${x}^{2} + 4 x - {e}^{6} = 0$

$\frac{- 4 \pm \sqrt{{4}^{2} - 4 \left(1\right) \left({e}^{6}\right)}}{2 \left(1\right)}$

$\frac{- 4 \pm \sqrt{16 + 4 {e}^{6}}}{2}$

$\frac{- 4 \pm \sqrt{4 \left(4 + {e}^{6}\right)}}{2}$

$\frac{- 4 \pm \sqrt{4} \sqrt{4 + {e}^{6}}}{2}$

$\frac{- 4 \pm 2 \sqrt{4 + {e}^{6}}}{2}$

$- 2 \pm \sqrt{4 + {e}^{6}}$