# How do you solve lnx + ln(x^2 + 1) = 8?

Sep 2, 2015

x≈14.37

#### Explanation:

$\ln \left(x\right) + \ln \left({x}^{2} + 1\right) = 8 \implies$use:$\log \left(a\right) + \log \left(b\right) = \log \left(a b\right)$

$\ln \left[x \left({x}^{2} + 1\right)\right] = 8 \implies$expand inside the brackets:

$\ln \left({x}^{3} + x\right) = 8 \implies$if:$\ln \left(x\right) = y \iff x = {e}^{y} :$

${x}^{3} + x = {e}^{8} \implies$using an equation solver:

x≈14.37

The other two roots are not real, they are complex.