Minimum of the function $b\log_b x$

Question

Why the function $b\log_b x$ has its minimum at $b=e$?

How to explain this?

I'm asking because I can't understand why ternary base has more economy than binary: http://www.americanscientist.org/issues/pub/2001/6/third-base/99999

Answer 1

Write the function as $f(b) = b\log_b x = \dfrac{b\ln x}{\ln b}$.

The derivative is $f'(b) = \dfrac{\ln b \cdot \ln x - b\ln x \cdot \tfrac{1}{b}}{(\ln b)^2} = \dfrac{(\ln b - 1)\ln x}{(\ln b)^2}$.

If $b < e$, then $\ln b < 1$, and $f'(b) < 0$. If $b > e$, then $\ln b > 1$, and $f'(b) > 0$.

Hence, $f(b)$ is decreasing over $1 < b < e$ and increasing over $b > e$.

Thus, $f(b)$ attains a minimum at $b = e$.

EDIT: Note that this assumes that $x > 1$, and so $\ln x > 0$. If $\ln x < 0$, then $f(b)$ attains a maximum at $b = e$ instead of a minimum.