Proof of Logarithms

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April 9th 2011, 09:02 PM
masoug

Proof of Logarithms

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

Quote:

Prove the following:
$log x < x$ given $x > 0$

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

Thanks!
April 9th 2011, 09:05 PM
Prove It

You can prove this using derivatives...
April 9th 2011, 09:16 PM
masoug

Okay... Is it possible without derivatives?
April 9th 2011, 09:29 PM
Prove It

Why would you want a different method?

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

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

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

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

Therefore $\displaystyle \ln{x} < x$ for all $\displaystyle x > 0$ .
April 10th 2011, 05:46 PM
masoug

Okay, that makes sense... So we can use growth rates of functions to determine which is the largest in the end.
April 10th 2011, 06:28 PM
Krizalid

Use a contradiction: suppose not, that means $\ln x\ge x,$ that would imply $x\ge e^x,$ but this is obviously false since $x>0.$