Hi,

I was looking at a math book when I stumbled across a problem:

I am a little puzzled. How would I prove it? Using proof by counterexample?Quote:

Prove the following:

$\displaystyle log x < x $ given $\displaystyle x > 0$

Thanks!

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- Apr 9th 2011, 09:02 PMmasougProof of Logarithms
Hi,

I was looking at a math book when I stumbled across a problem:

Quote:

Prove the following:

$\displaystyle log x < x $ given $\displaystyle x > 0$

Thanks! - Apr 9th 2011, 09:05 PMProve It
You can prove this using derivatives...

- Apr 9th 2011, 09:16 PMmasoug
Okay... Is it possible without derivatives?

- Apr 9th 2011, 09:29 PMProve It
Why would you want a different method?

First note that $\displaystyle \displaystyle \ln{x} < 0$ for all $\displaystyle \displaystyle x \in (0,1)$, while $\displaystyle \displaystyle x > 0$ for all $\displaystyle \displaystyle x > 0$. So $\displaystyle \displaystyle \ln{x} < x$ for all $\displaystyle \displaystyle x \in (0, 1)$.

Then note that $\displaystyle \displaystyle \frac{d}{dx}(\ln{x}) = \frac{1}{x}$ and $\displaystyle \displaystyle \frac{d}{dx}(x) = 1$.

For $\displaystyle \displaystyle x \geq 1$, we have $\displaystyle \displaystyle \frac{1}{x} < 1$

So that means $\displaystyle \displaystyle \ln{x}$ grows slower than $\displaystyle \displaystyle x$ does, so can never catch up.

Therefore $\displaystyle \displaystyle \ln{x} < x$ for all $\displaystyle \displaystyle x > 0$. - Apr 10th 2011, 05:46 PMmasoug
Okay, that makes sense... So we can use growth rates of functions to determine which is the largest in the end.

- Apr 10th 2011, 06:28 PMKrizalid
Use a contradiction: suppose not, that means $\displaystyle \ln x\ge x,$ that would imply $\displaystyle x\ge e^x,$ but this is obviously false since $\displaystyle x>0.$