Value of b for b by Logarithm Base b of x to be Minimum

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Theorem

Let $x \in \R_{> 0}$ be a (strictly) positive real number.

Consider the real function $f: \R_{> 0} \to \R$ defined as:

$f \left({b}\right) := b \log_b x$


$f$ attains a minimum when

$b = e$

where $e$ is Euler's number.


Proof

From Derivative at Maximum or Minimum, when $f$ is at a minimum, its derivative $\dfrac \d {\d b} f$ will be zero.

Let $y = f \left({b}\right)$.

We have:


\(\displaystyle y\) \(=\) \(\displaystyle b \log_b x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {b \ln x} {\ln b}\) Change of Base of Logarithm
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\d y} {\d b}\) \(=\) \(\displaystyle \frac {\ln b \frac \d {\d b} \left({b \ln x}\right) - b \ln x \frac \d {\d b} \ln b} {\left({\ln b}\right)^2}\) Quotient Rule for Derivatives
\(\displaystyle \) \(=\) \(\displaystyle \frac {\ln b \ln x - b \ln x \frac 1 b} {\left({\ln b}\right)^2}\) Derivative of Natural Logarithm, Derivative of Identity Function
\(\displaystyle \) \(=\) \(\displaystyle \frac {\ln x} {\ln b} \left({1 - \frac 1 {\ln b} }\right)\) simplifying
\(\displaystyle \) \(=\) \(\displaystyle \frac {\ln x} {\ln b^2} \left({\ln b - 1}\right)\) simplifying

Thus:

\(\displaystyle \dfrac {\d y} {\d b}\) \(=\) \(\displaystyle 0\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\ln x} {\ln b}\) \(=\) \(\displaystyle \frac {\ln x} {\left({\ln b}\right)^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \ln b\) \(=\) \(\displaystyle 1\) simplifying
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(=\) \(\displaystyle e\) Definition of Natural Logarithm


To determine that $f$ is a minimum at this point, we differentiate again with respect to $b$:

\(\displaystyle \frac {\d^2 y} {\d b^2}\) \(=\) \(\displaystyle \frac \d {\d b} \left({\frac {\ln x} {\ln b^2} \left({\ln b - 1}\right) }\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\ln x} b \left({\frac {\ln b - 2 \left({\ln b - 1}\right)} {\left({\ln b}\right)^3} }\right)\)

Setting $b = e$ gives:

$\dfrac {\d^2 y} {\d b^2} \Bigg \vert_{b \mathop = e} = \dfrac {\ln x} e \dfrac {\left({1 - 2 \left({0}\right)}\right)} 1$

which works out to be (strictly) positive.

From Second Derivative of Strictly Convex Real Function is Strictly Positive, $f$ is strictly convex at this point.

Thus $f$ is a minimum.

$\blacksquare$


Sources