# How do you solve lnx+ln(x+2)=4?

Dec 15, 2015

$x = - 1 + \sqrt{{e}^{4} + 1}$

#### Explanation:

Use the following logarithm rule: ${\log}_{a} b + {\log}_{a} c = {\log}_{a} \left(b c\right)$

$\ln \left(x \left(x + 2\right)\right) = 4$

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

Recall that $\ln x = {\log}_{e} x$.

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

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

${x}^{2} + 2 x + 1 = {e}^{4} + 1$

${\left(x + 1\right)}^{2} = {e}^{4} + 1$

$x + 1 = \pm \sqrt{{e}^{4} + 1}$

$x = - 1 \pm \sqrt{{e}^{4} + 1}$

Plug in both values for $x$. Notice that only the positive since version works since it's impossible to take the logarithm of a negative number.

$x = - 1 + \sqrt{{e}^{4} + 1} \approx 6.456$

graph{ln(x)+ln(x+2)-4 [-7.54, 20.94, -6.05, 8.19]}