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?Prove the following:
$\displaystyle log x < x $ given $\displaystyle x > 0$
Thanks!
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 $\displaystyle \displaystyle x > 0$.