*Note: the logs are with base 2 (computer scince question).
Let
$$ f(n) = \max\{\sqrt{n^3}\lg n, \sqrt[3]{n^4}\lg^5n \} = \max\{f_1(n),f_2(n)\} $$
I want to find which function is the max between the two functions such that the bigger will satisfies:
$$ \exists n_0 \in N, \forall n > n_0: f_1(n) > f_2(n) $$
What i tried:
I tried to start with comparing the two functions, hopinf to find the cross point, than just checking what happend after and before the cross point.
Therefore, i try to solve:
$$ \sqrt{n^3}\lg n =\sqrt[3]{n^4}\lg^5n $$
Solving this i got to: $$ \sqrt[3]{n} = \lg^4n $$
Maybe i miss something, but i dong know how to solve this...
I would like a hint more than a solution.
Thank you very much for the answers.
Let's play around with real $x$ to see what we get. Rearranging the equation we have
$$x^{\frac{1}{6}} = \lg^4(x) \implies 2^{x^{\frac{1}{24}}} = x$$
Then move everything to one side and with some rearrangement we get that
$$-\frac{\log 2}{24} = -\frac{\log 2}{24}x^{\frac{1}{24}}\exp\left(-\frac{\log 2}{24}x^{\frac{1}{24}}\right) \implies x = \frac{24^{24}}{\log^{24} 2}W^{24}\left(-\frac{\log 2}{24}\right) \approx 2.04$$
but also we have the two expressions are equal when $x=1$ (a solution we lost when we divided both of the original expressions by $\lg x$). If you would only like a hint stop here. Otherwise I continue the solution below.
Since these are the only places the expressions are equal, plug in a nice value of $x\geq 3$, say $x=2^6$:
$$\sqrt{\left(2^6\right)^3}\lg(2^6) = 3\times 2^{10}$$
$$\sqrt[3]{\left(2^6\right)^4}\lg^5(2^6) = 3^5 \times 2^{13}$$
so clearly for $ n > 2$ $$ \sqrt[3]{n^4}\lg^5(n) > \sqrt{n^3}\lg(n) $$