Upper Bound of Natural Logarithm

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.


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

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