# Upper Bound of Natural Logarithm

From ProofWiki

## Theorem

Let $\ln y$ be the natural logarithm of $y$ where $y \in \R, y > 0$.

Then:

- $\ln y \le y - 1$
- $\forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$

## Proof

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

We have that the natural logarithm function is concave.

From Mean Value of Convex and Concave Functions, we have:

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

From Derivative of Natural Logarithm we have:

- $D \ln 1 = \dfrac 1 1 = 1$

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 $\ln x \le \dfrac {x^s} s$:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle s \ln x\) | \(=\) | \(\displaystyle \) | \(\displaystyle \ln {x^s}\) | \(\displaystyle \) | \(\displaystyle \) | Logarithms of Powers | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \) | \(\displaystyle x^s - 1\) | \(\displaystyle \) | \(\displaystyle \) | from above | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \) | \(\displaystyle x^s\) | \(\displaystyle \) | \(\displaystyle \) |

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

$\blacksquare$

## Sources

- K.G. Binmore:
*Mathematical Analysis: A Straightforward Approach*(1977)... (previous)... (next): $\S 14.3 \ (2)$