# For the function $y=\ln(x)/x$: Show that maximum value of y occurs when $x = e\ldots$

For the function $$y=\ln(x)/x$$:

Show that maximum value of $$y$$ occurs when $$x = e$$.

Using this information, show that $$x^e for all positive values of $$x$$.

Two positive integers, $$a$$ and $$b$$, where $$a < b$$, satisfy the equation $$a^b = b^a$$. Find $$a$$ and $$b$$ , and show that these are unique solutions.

For the first probelm, I was thinking of getting the derivative of the function (which is $$(\ln(x)+1)/x^2)$$ and using sign charts in order to show the max.

But I'm not sure on how I would use that for the second problem nor the third problem.

Thanks!

• Either the inequality on the third line should read $x^e \leqslant e^x$, or else it should be stated that $x \ne e$. – Calum Gilhooley Oct 23 '18 at 20:33

Let $$f(x)=\frac{\ln(x)}{x}$$

for $$x>0$$.

$$f'(x)=\frac{1-\ln(x)}{x^2}$$

$$f'(x)=0\iff \ln(x)=1\iff x=e.$$

thus

$$(\forall x>0) \;\; \frac{\ln(x)}{x}\le f(e)$$ or $$(\forall x>0) \;\; e\ln(x)\le x$$

and $$(\forall x>0)\;\; \ln(x^e)\le \ln(e^x)$$

For the other

$$a^b=b^a \implies b\ln(a)=a\ln(b)$$

$$\implies f(a)=f(b)$$

$$e\approx 2.8 \implies a\in\{0,1,2\}$$

$$\implies a=2\;\; b=4$$

You're on the right track. Let $$y(x)=\frac{\log(x)}{x}$$.

Differentiating we find $$y'(x)=\frac{1-\log(x)}{x^2}$$ whence $$y'(x)=0$$ when $$x=e$$. It is easy to see that this local extremum is the maximum. Hence

$$\frac{\log(x)}{x}\le \frac1e\tag1$$

Rearranging $$(1)$$ we find that

$$x^e\le e^x$$

as was to be shown.

For the third part, note that if $$a^b=b^a$$ then

$$\frac{\log(a)}{a}=\frac{\log(b)}{b}\tag 2$$

We know that the function $$y(x)=\frac{\log(x)}{x}$$ has a maximum at $$x=e$$ and that $$y(x)$$ is concave. Hence if $$(2)$$ has a solution for integer values of $$a$$ and $$b$$, then $$b$$ must be less than $$e<3$$ and $$a\ge 3$$. The only possible solution is $$b=2$$ in which case $$a=4$$. And we are done.

HINT

Let consider

$$f(x)=\frac{\ln x}{x}\implies f'(x)=\frac{1-\log x}{x^2}=0\implies \ldots$$

and then note

$$x^e

take the derivative w.r.t. $$x$$ and equate it to $$0$$ $$\frac{d}{dx} \frac{ln(x)}x=0$$ $$\frac{1-ln(x)}{x^2}=0$$ $$1=ln(x)$$ $$x=e$$

$$x^e for every $$x>0$$

$$ln(x^e)

$$e*ln(x)

$$\frac{ln(x)}x<\frac{1}e$$ (the sign of the inequality does not change since $$x>0$$)

We have shown that the maximum value of $$\frac{ln(x)}x$$ is attained at $$x=e$$ and hence the inequality is satisfied.

for the third question I suggest you to read this post: https://mathoverflow.net/questions/22230/ab-ba-when-a-is-not-equal-to-b

1) Yes. Use derivatives and you will get it. So $$\frac {\ln x}{x} \le \frac {\ln e}{e} = \frac 1 e$$

2) for $$x > 0$$; $$x^e \le e^x \iff \ln x^e \le \ln e^x \iff e*\ln x \le x\iff \frac {\ln x}x \le \frac 1e$$

3) The thing is that by derivative you will note that for $$x < e$$ then $$\frac {\ln x}x$$ is increasing but of $$x > e$$ then $$\frac {\ln x}x$$ is decreasing so if $$\frac {\ln x} x = \frac {\ln w } w= k$$ and $$x < w$$ then $$x < e < w$$ and $$x, w$$ are unique.

If $$a^b = b^a$$ then $$b \ln a = a \ln b$$ and $$\frac {\ln a}{a} = \frac {\ln b} b$$ and $$a < b$$ so $$a < e < b$$. So $$a = 1,2$$. And for each $$a$$ any $$b$$ (if any) so that $$\frac {\ln b}{b} = \frac {\ln a}{a}$$ and $$b > a$$ will be a unique.

Here's another approach (to the first part). We have $$\ln(x)/x \leqslant 1/e$$ if and only if $$\ln(x) \leqslant x/e$$, i.e. $$1 + t \leqslant e^t$$, where $$t = \ln(x/e)$$. As shown in several ways here or here (take your pick!), this inequality holds for all real $$t$$.

• Here's another thread that proves essentially the same inequality. It should probably be an faq. – Calum Gilhooley Oct 23 '18 at 21:29