# Upper Bound of Natural Logarithm

## Theorem

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

Then:

$(1): \quad \ln y \le y - 1$
$(2): \quad \forall s \in \R_{>0}: \ln x \le \dfrac {x^s} s$

## Proof

First, to show that $\ln y \le y - 1$:

$\ln y - \ln 1 \le \left({D \ln 1}\right) \left({y - 1}\right)$
$D \ln 1 = \dfrac 1 1 = 1$

So:

$\ln y - \ln 1 \le \left({y - 1}\right)$

But from Logarithm of 1 is 0:

$\ln 1 = 0$

Hence the result.

$\Box$

Next, to show that $\ln x \le \dfrac {x^s} s$:

 $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle s \ln x$$ $$=$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle \ln {x^s}$$ $$\displaystyle$$ $$\displaystyle$$ Logarithm of Power $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\le$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle x^s - 1$$ $$\displaystyle$$ $$\displaystyle$$ from above $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\displaystyle$$ $$\le$$ $$\displaystyle$$  $$\displaystyle$$ $$\displaystyle x^s$$ $$\displaystyle$$ $$\displaystyle$$

The result follows by dividing both sides by $s$.

$\blacksquare$