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.

$\blacksquare$

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle s \ln x\)	\(=\)	\(\displaystyle \ln {x^s}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Logarithms of Powers
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle x^s - 1\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle x^s\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

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

$\blacksquare$

Sources

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