Prove $\log(x)^{10} < x$ (for $x > 10^{10}$)

Question

I need to prove that $\log(x)^{10} < x$ for $\ x>10^{10}$ It's pretty clearly true to me, but I need a good proof of it. I tried induction, and got stuck there.

Answer 1

You want, for $x > 10^{10}$, $(\log(x))^{10} < x $. Setting $x = y^{10}$, this is $(\log(y^{10}))^{10} < y^{10} $ for $y > 10$ or $\log(y^{10}) < y$ or $y^{10} < 10^y$.

Since $x^{1/x}$ is decreasing for $x > e$, if $y > 10$, $10^{1/10} > y^{1/y}$ or $10^y > y^{10}$.

Answer 2

Assuming this is $\log$ base $10$, you know that $\log_{10}(10^n)^{10}=n^{10}$ (this follows from the identity $\log_{a}(a^{k})=k$), so it seems you would have to verify that $10^{n}>n^{10}$ for $n>10$. I am not sure that this makes it easier for you, but atleast the log is gone. You can then show that the derivative of $n^{10}$ grows slower than $10^n$ (My rep is not high enough to leave this as a comment).