Maximum value of $\log(n)/n$ where $n$ is a natural number?

Question

I was trying to find the maximum value of $\log(n) / n$ for the case of natural numbers. I could think only of graphing it so here is the graph-

It is seen that maximum occurs at $n =3$ with maximum value $\log(3)/3$ but why is it so? any analytical technique like differentiating tit and equating it to zero ?

Now I thought of generalizing it to positive real numbers like the question can be re-framed as what is the maximum value of $\log(x) / x$ where $x$ is a positive real number now I tried this by differentiating it w.r.t $x$

$$\frac{d}{dx}\left(\frac{\log(x)}{x}\right) = 0$$

$$\frac{1}{x^2} [x\cdot \frac{1}{x} - \log(x)\cdot 1] = 0$$

$$\log(x) = 1$$ and if it is in base $10$ then $x = 10$ now is this correct ?

Any help is great!.

Answer 1

Your approach was correct, but there was a small mistake.

$$\frac{d}{dx}\left(\frac{\log_a(x)}{x}\right) = 0$$

$$\frac{1}{x^2} [x.\frac{1}{x} \log_a (e) - \log_a(x).1] = 0$$

$$ \log_a(x) = \log_a(e) \implies x = e$$

The solution of your problem will be either $n = 2$ or $3$.