Upper Bound of Natural Logarithm

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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 $s$.

$\blacksquare$


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