How do you solve ln(x^2+12)=lnx+ln8?

Oct 17, 2016

$x = 2$ and $x = 6$

Explanation:

$\ln \left({x}^{2} + 12\right) = \ln x + \ln 8$
We can rewrite this as $\ln \left({x}^{2} + 12\right) = \ln 8 x$
Removing the ln on both sides,give
${x}^{2} + 12 = 8 x$
Rearranging
${x}^{2} - 8 x + 12 = 0$
$\left(x - 6\right) \left(x - 2\right) = 0$
so x=6 or x=2
We have to check the conditions for ln
${x}^{2} + 12 > 0$and $8 x > 0$
So the solution is ok