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

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.

• Well, it looks like you want to prove for $x \in \mathbf{R}$ right? So you cannot use induction. Commented Jan 29, 2016 at 1:27
• You need to prove $x^{10}<10^x$ which isnt true i think Commented Jan 29, 2016 at 1:29
• I just need to prove it for x>10^10, and it is true. I don't need an induction proof, just any proof Commented Jan 29, 2016 at 1:35
• I can use any means necessary as long as it's sufficient proof. How would I go about using derivatives in this case? Commented Jan 29, 2016 at 1:40
• The inequality is equivalent to $\log_{10}(x) < x^{1/10}$ which by taking $z = x^{1/10}$ is seen to be equivalent to $10\log_{10}(z) < z$ for all $z>10$. Here you can for example take $f(z) = z - 10\log_{10}(z)$ and show that $f'(z) > 0$ for $z>10$ togeather with $f(10) = 0$ to get the desired conclusion. Commented Jan 29, 2016 at 1:53

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).
• Alternatively $10^n > n^{10} \iff 10^{1/10}>n^{1/n}$ so it is enough to show $t^{1/t}$ has its only maximum when $t=e$, which can easily be shown. +1. Commented Jan 29, 2016 at 2:13
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 $10^y > y^{10}$.