Can we minimize $m\log_m{n}$, given $n$?

Question

If we are given $n$, a positive real, can we find a the positive real $m$ that minimizes the function:

$$m\log_m{n}$$

I'd prefer to find the function that gives a value for $m$, but I'm also interested in asymptotic bounds for $m$.

This is similar to my question here.

WHAT I HAVE

I start with

$$x = m\log_m{n}$$ $$x = m\frac{\log n}{\log m}$$

Then, since $\log n$ is constant, we simply want to minimize $$\frac{m}{\log m}$$

Is this correct? I'm really hoping that someone can solve the original equation. It's not homework.

If $n>1$ it is correct. – Pocho la pantera Sep 03 '13 at 01:52 — Pocho la pantera, Sep 03 '13 at 01:52
@Pocholapantera: We do have $n>1$. – Matt Groff Sep 03 '13 at 01:53 — Matt Groff, Sep 03 '13 at 01:53

score 4 · Accepted Answer · answered Sep 03 '13 at 02:05

4

Yes, you're correct so far. Let $f(m)=\frac{m}{\log m}$; then

$$f'(m) = \frac{\log m - 1}{(\log m)^2} = 0 \iff m = e$$

Hence, $f$ has an extreme point at $e$. This is easily shown to be a global minimum.

answered Sep 03 '13 at 02:05

Actually there is no global minimum since $\lim_{x \to 1^{-}} f(x) = -\infty$. – Pratyush Sarkar Sep 03 '13 at 02:17

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