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

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:

Thus:

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

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.