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
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $29$