# 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 $\dfrac \d {\d b} f$ 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.

$\blacksquare$