# Proof of Logarithms

• Apr 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:
$\displaystyle log x < x$ given $\displaystyle x > 0$
I am a little puzzled. How would I prove it? Using proof by counterexample?

Thanks!
• Apr 9th 2011, 09:05 PM
Prove It
You can prove this using derivatives...
• Apr 9th 2011, 09:16 PM
masoug
Okay... Is it possible without derivatives?
• Apr 9th 2011, 09:29 PM
Prove 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 PM
masoug
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 PM
Krizalid
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 $\displaystyle x>0.$