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

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

Using this information, show that $x^e <e^x$ 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!

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

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

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

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

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

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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.

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HINT

Let consider

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

and then note

$$x^e <e^x \iff \ln (x^e) < \ln (e^x)$$

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


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

$ln(x^e)<ln(e^x)$

$e*ln(x)<x$

$$\frac{ln(x)}x<\frac{1}e$$ (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

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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.

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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 $t$.

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

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