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:

$\map f b := b \log_b x$


$f$ 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 $y = \map f b$.

We have:


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

Thus:

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


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

\(\ds \frac {\d^2 y} {\d b^2}\) \(=\) \(\ds \frac \d {\d b} \paren {\frac {\ln x} {\ln b^2} \paren {\ln b - 1} }\)
\(\ds \) \(=\) \(\ds \frac {\ln x} b \paren {\frac {\ln b - 2 \paren {\ln b - 1} } {\paren {\ln b}^3} }\)

Setting $b = e$ gives:

$\valueat {\dfrac {\d^2 y} {\d b^2} } {b \mathop = e} = \dfrac {\ln x} e \dfrac {\paren {1 - 2 \paren 0} } 1$

which works out to be (strictly) positive.

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

Thus $f$ is a minimum.

$\blacksquare$


Sources