Master theorem with $T(n) = 6T(n/10) + \lg^2 n$

Question

So I think this falls under case 1...but I am not sure how to formally prove that $n^{lg_{10} 6}$ is asymptotically greater than $\lg^2 n$. I tried using L'hopitals but kinda reached a dead end. Here's what i have so far

$$\lim_{n \rightarrow \infty} \frac{d(\lg^2 n)}{d(n^{\lg_{10}6})} = \frac{2 \lg n (\frac{1}{n})}{(\lg_{10}6) n ^{(\lg_{10}6) - 1}}$$

Is there some better way I can think about proving case 1?

Answer 1

Yuval Filmus pretty much showed the way in the comment above. I will just elaborate on his approach.

First, we raise both to the power lg₆10.

(n^lg₁₀6)^lg₆10 = n^{(lg₁₀6 x lg₆10)} = n

and,

(lg²n)^lg₆10 = (lg n)^{2 x lg₆10} = (lg n)^2.57

Now, taking log of both of these, we get lg(n) and 2.57xlg(lg(n)) respectively.

lg(n) is asymptotically greater than lg(lg(n)), and thus, our case is proved.