# How do you solve log_10(a^2-6)>log_10a?

Nov 10, 2016

graph{(y-ln(x^2-6)/ln10)(y-ln(x)/ln10)=0 [-1.785, 10.7, -1.773, 4.467]}

$a > 3$

#### Explanation:

First ensure the logs exists:

$\left\{\begin{matrix}{a}^{2} - 6 > 0 \\ a > 0\end{matrix}\right. \implies a > \sqrt{6}$

Then cosider that ${\log}_{10}$ is a growing function so the given relation is equivalent to

$\left\{\begin{matrix}a > \sqrt{6} \\ {a}^{2} - 6 > a\end{matrix}\right.$

$\left\{\begin{matrix}a > \sqrt{6} \\ {a}^{2} - a - 6 > 0\end{matrix}\right.$

$\left\{\begin{matrix}a > \sqrt{6} \\ \left(a - 3\right) \left(a + 2\right) > 0\end{matrix}\right. \implies a > 3$