# Upper Bound of Natural Logarithm/Proof 1

## Theorem

Let $\ln x$ be the natural logarithm of $x$ where $x \in \R_{>0}$.

Then:

$\ln x \le x - 1$

## Proof

$\ln$ is (strictly) concave.
$\ln x - \ln 1 \le \paren {\dfrac \d {\d x} \ln 1} \paren {x - 1}$
$\dfrac \d {\d x} \ln 1 = \dfrac 1 1 = 1$

So:

$\ln x - \ln 1 \le \paren {x - 1}$

But from Logarithm of 1 is 0:

$\ln 1 = 0$

Hence the result.

$\blacksquare$