Prove that $\exists x\in \mathbb{R}: x > 2 \land 10^x> x^{10}$

Question

$$\exists x\in \mathbb{R}: x > 2 \land 10^x> x^{10}$$

I tried:

$$x>2 \land 10^x>x^{10} = x > 2 \land 10^2>2^{10} = x > 2 \land 100 > 1024$$

Honestly I am not entirely sure of what the expression means. I think it means "There is an $x$ in $\mathbb{R}$ that is greater than 2 so that $10^x > x^{10}$.

Is this correct? How do I prove this?

Answer 1

Here's one. Take $x=2^{10}$. We have $$10^x=10^{1024}=(10^4)^{256}>x^{256}>x^{10}.$$

Answer 2

Let $\log x$ denote $\log_{10}x$. Note that $$ 10^x>x^{10}\ \Leftrightarrow\ \log(10^x)>\log(x^{10}) \ \Leftrightarrow\ x\overbrace{\log10}^{=1}>10\log x \ \Leftrightarrow\ \frac{x}{\log x}>10 $$ Taking $x=100$ yields $x>2$ and $$ \frac{x}{\log x}=\frac{100}{\log 100}=\frac{100}{2}=50>10 $$ as required.

Answer 3

Obviously, $10^{10}\ge10^{10}$, then $10^{10.00001}>10.00001^{10}$, as the exponential grows faster than the power.

Answer 4

Okay, Thought process:

$x=2; 10^2 = 100 < 2^{10} = 1024$

$x=3; 10^3 = 1000 < 3^{10} > 1024$ and this is getting nowhere.

But $x = 10; 10^{10} = 10^{10}$. So maybe $10^{11} > 11^{10}$. And it is.

A single example is a proof. But...

Okay a real proof.

Let $f(x) = 10^x - x^{10}$. $f(10) = 10^{10}-10^{10} = 0$.

$f'(x) = 10^x*\ln(x) - 10*x^9$. $f'(10) = 10^{10}*\ln{10} - 10*10^9 = 10^{10}(\ln{10} - 1)$.

Well $10 > e$ so $\ln{10} > 1$ so $f'(10) > 0$.

So $f$ is increasing at $x = 10$. So there exists an $x > 10$ where $f(x) > f(10)$.

And that's that, isn't it?

$\exists x > 10 > 2| 10^x - x^{10} > 0$.

Answer 5

Well, the simplest way to prove the existence of something is to explicitly provide it.

We have: $11 > 2 \land 10^{11} = 100\,000\,000\,000 > 25\,937\,424\,601 = 11^{10}$

So how did I get to $11$? Well, obviously, for $x=10$, we have $10^x=x^{10}$, so something larger than $10$ should be a good candidate. I then tried $11$ and it worked.

And yes, showing that $11$ fulfils the condition is a perfectly valid proof for the existence of an $x$ that fulfils the condition.