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

$f$ 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 $\dfrac \rd {\rd b} f$ will be zero.

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

We have:

Thus:

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

Setting $b = e$ gives:
 * $\dfrac {\rd^2 y} {\rd b^2} \Bigg \vert_{b \mathop = e} = \dfrac {\ln x} e \dfrac {\left({1 - 2 \left({0}\right)}\right)} 1$

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.