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

Jul 3, 2016

Use the log rule ${\log}_{a} \left(m\right) + {\log}_{a} \left(n\right) = {\log}_{a} \left(m \times n\right)$

#### Explanation:

ln(x(x + 2) = 1

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

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

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

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

x = (-2 +- sqrt(2^2 - 4(1)(-e)))/(2(1)

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

$x = 0.93 \mathmr{and} - 2.93$

However, checking the solutions in the original equation we find that $x = - 2.93$ is extraneous.

Hence, the solution set is $\left\{0.93\right\}$.

Hopefully this helps!