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Suppose $x$ and $y$ are two positive numbers satisfying the equation $x^y = y^x .$ Which of the following are true?

(a) For all $x > 1,$ there always exist a $y > x$ such that the above equation holds.

(b) For all $x > e$ there is always a $y > x$ such that the above equation holds.

(c) For all $1 < x < e$ there is always a $y > x$ such that the above equation holds.

(d) If $x < 1,$ the y must be equal to $x.$

I was struggling with this problem for a long time. I was primarily focussing upon the idea, that $x^y = y^x$ has solutions $(x,y)=((1+\frac 1t)^t,(1+\frac 1t)^{t+1}),$ where $t\in\Bbb R^*.$ Using this,I tried hovering over the options given.

EDIT: As mentioned by @Macavity, this approach stands invalid, as the parametric solution, I considered here, is not covering all the solutions of the given equation (if I understand him correctly). I proceeded following his suggestions by considering the graph of $\frac {\ln x}{x}$. To have a look at it please proceed further and my new attemt is marked with an italics case to make it easier to comprehend.

We first consider the case,when $x=y$, then, $(1+\frac 1t)^t=(1+\frac 1t)^{t+1}$. Assuming, if $(1+\frac 1t)\neq 0$, then, $$1=(1+\frac 1t)\implies \frac 1t=0,$$which is absurd. If, $$(1+\frac 1t)=0,$$ then, $t=-1.$ We observe, $x=y$, only if, $t=-1$, which implies $x=0,y=0.$ So, we can safely conclude the option D is incorrect.

Now, we try to determine, the case, when $x>y$ i.e this is possible only when $$(1+\frac 1t)^t>(1+\frac 1t)^{t+1}\implies 1>(1+\frac 1t)$$, this is true, under the assumption, $(1+\frac 1t)^t>0.$ Now, $$(1+\frac 1t)<1\implies \frac 1t<0\implies t<0.$$ So, we conclude, $x>y$ if $t<0.$ Similarly, we say, $x<y$ only if, $t>0.$ But, if $t>0$, then, $x,y>0.$ So, we can say, option $A,B$ and $C$ are observe to hold good with this derived result. Hence, the answer is options $A,B$ and $C.$

However, the answer is given as option $C$ and $D.$ I don't quite find any error in my approach. I am confused about this. I am unsure, whether I missed something basic or not.

My second attemt, following @Macavity's suggestion, goes like this:

I tried sketching the graph of $y=\frac{\ln x}{x}$. It matched with this,

enter image description here

Then, I proceeded with a simplication of the question. Any solution, of the equation, $\frac{\ln x}{x} =\frac{\ln y}{y}$ is also a solution of the equation, $x^y = y^x.$ But the converse might not be true. Then, I again started hovering over the given options, and I started with option $A$:

Option $A$ states, that, " For all $x > 1,$ there always exist a $y > x$ such that the above equation holds." Now, if this condition satisfies, then it means, for each $x>1$ we will find a $y$ such that the ordered pair $(x,y)$ is a solution of the given equation, and also of the equation, $\frac{\ln x}{x} =\frac{\ln y}{y}$, as $x>1.$ Now, from the graph of the function $y=\frac{\ln x}{x}$ it can be easily verified, that this is not the case since, the function $\frac{\ln x}{x}$ is strictly increasing when $1<x<e$ and strictly decreasing when $x\geq e$. So, $Option A$ is incorrect.

Next, we try to verify the validity of option $B$:

In option $B$ it is presented, that "For all $x > e$ there is always a $y > x$ such that the above equation holds." Now, similar to the above reasonining we may conclude, $B$ is incorrect. This is because, if this condition satisfies, then it means, for each $x>e$ we will find a $y$ such that the ordered pair $(x,y)$ is a solution of the given equation, and also of the equation, $\frac{\ln x}{x} =\frac{\ln y}{y}$, as $x>e.$ Now, from the graph of the function $y=\frac{\ln x}{x}$ it can be easily verified, that this is not the case since, the function $\frac{\ln x}{x}$ is strictly decreasing when $x>e.$ So, $Option B$ is incorrect.

Now, we proceed to check the validity of Option $C$ :

In option $C$ it's asserted that " For all $1 < x < e$ there is always a $y > x$ such that the above equation holds." We proceed like the previous two cases. If this condition satisfies, then it means, for each $1<x<e$ we will find a $y$ such that the ordered pair $(x,y)$ is a solution of the given equation, and also of the equation, $\frac{\ln x}{x} =\frac{\ln y}{y}$, as $x>1.$ Now, from the graph of the function $y=\frac{\ln x}{x}$ it can be easily verified, that this is the case since, the function $\frac{\ln x}{x}$ is strictly decreases when $x>e,$ and as seen on the graph, the $y$ co-ordinate again takes all the values that were taken previously when $x$ was in the range $1<x<e.$ So, $Option C$ is correct.

Now, we proceed to validate, $Option D$ :

In Option $D$ it is asserted that "If $x < 1,$ the y must be equal to $x.$" We proceed like the other cases. If this condition satisfies, then it means, for each $x<1$ we will find a $y=x,$ such that the ordered pair $(x,y)\equiv (x,x)$ is a solution of the given equation, and also of the equation, $\frac{\ln x}{x} =\frac{\ln x}{x}$, if $0<x<1.$ Now, from the graph of the function $y=\frac{\ln x}{x}$ it can be easily verified, that this is the case,if $0<x<1$ since, the function $\frac{\ln x}{x}$ is strictly increasing when $0<x<1.$ So, $Option C$ might be correct. We can't argue about about the case, when $x<0$ as $y$ is not defined there. So, we can't be sure of Option $D$. Nevertheless, this is the only option left and the question states Only two options are correct. Hence, $Option D$ should be the correct option.

I feel the issue is now resolved, my answers are now matching with the given one. I think my approach, now is devoid of any more loopholes. Thanks to @Macavity, for the useful suggestions. Indeed, graphs are nice little ways, to solve these sorts of problems.

I saw, or realize (rather,) that some users don't advocate the usage of images in this site. But I couldn't help it, as without the help of the image of the graph, I couldn't quite explain myself or my reasonings sufficiently and optimally.

Arthur
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    For your first case, obviously $(x,y)=(t,t)$ is always a solution. The parameterisation you used ignores this case, hence your conclusion for option D is wrong. For this problem, it is much easier if you approach it by studying the function $t^{1/t}$ or if you prefer, $\log(t)/t$. – Macavity Apr 29 '23 at 09:59
  • Not really. In the other cases, you have used a property of the parameters to conclude there is no restriction. However the logic is similar to saying $P \implies Q$ which is true, therefore $P$ is true. Such a conclusion isn't valid. Just because the parameters allow a solution of form $x>y$, doesn't mean the specific conditions mentioned in the options $A, B$ are being met. In fact $A, B$ are not correct options, for e.g. for option $A$, take $x = 2e$, can you find a $y>x$ that works? – Macavity Apr 30 '23 at 06:35
  • @Macavity I have edite the original post. I hope this is what you meant as I explained there? – Arthur Apr 30 '23 at 07:31
  • Other discussions of $x^y=y^x$: https://math.stackexchange.com/questions/944326/find-the-all-possible-real-solutions-of-xy-yx and https://math.stackexchange.com/questions/9505/xy-yx-for-integers-x-and-y and https://math.stackexchange.com/questions/2264576/is-there-only-one-pair-of-unequal-rational-numbers-with-nm-mn and https://math.stackexchange.com/questions/199235/determine-the-number-of-solutions-of-the-equation-nm-mn and https://math.stackexchange.com/questions/199235/determine-the-number-of-solutions-of-the-equation-nm-mn? and probably many, many more. – Gerry Myerson Apr 30 '23 at 07:45
  • @GerryMyerson Yes, using these all links, I constructed the parametric solution for $x^y=y^x$ as done in my first attempt. But as, Macavity points, out, (if I understand him correctly) that the parametric representation of the solution don't quite cover all the cases, possible. So, using that, is not very much benificial for solving this problem, as presented. If I am not wrong, all these links point out or revolves about the parametric solution that I used and do not answer this question of mine as it seems. – Arthur Apr 30 '23 at 07:52
  • One solution to $y^x = x^y$ is $x = 2$ and $y = 4.$ While this is no way to go about a proof, as far as answering multiple-choice questions it is a great place to start. – user317176 Apr 30 '23 at 08:27
  • The graph isn't a rigorous mathematical proof, at best it can illustrate things. If you know calculus, you can establish that the function is increasing in the interval $(0, e)$ and decreasing for $x>e$. In fact the image of $(1, e)$ and $(e, \infty)$ are identical. Hence $C$ is a correct option, for any $x \in (1, e)$, we can find a $y>e$ s.t. $f(x)=f(y)$ or equivalently $x^y=y^x$. – Macavity Apr 30 '23 at 09:46
  • @Macavity I could establish the fact that "the function is increasing in the interval $(0,e)$ and decreasing for $x>e.$ But I dont understand what you mean by the phrase "the image of $(1,e)$ and $(e,\infty)$ are identical. " ? Also, correct me if I am wrong as I think the function you are talking about in your comment is $frac{\ln x}{x}.$ If this is the case, then also, I am not entirely convinced about the validity of Option $C$ from the graph in my original post, since, the graph is first increasing at the same rate upto $e$ and then starts decreasing after $e$ at the same rate but – Arthur May 01 '23 at 11:25
  • @Macavity only for a finite amount of time. After a certain time, or rather after a particular x-coordinate the intensity of decreasing don't remain the same as the intensity of decreasing. The intensities become very much different due to which, their images after $e$ dont quite match with the images before $e$ indefinitely. It would be helpful if you clarify what you mean ? It might be the case that I am missing something. – Arthur May 01 '23 at 11:27
  • @FdstZfsy Whichever $f$ you take, note it is continuous in $(1, \infty)$. Further $f(1)=\lim_{x\to \infty} f(x)$. Thus $f$ maps $(1, e)$ and $(e, \infty)$ both into $(f(1), f(e))$. The "intensity" of decrease doesn't matter, which is why the calculations help become precise as opposed to the graph. – Macavity May 01 '23 at 12:31
  • @Macavity I am now, totally convinced that $C$ is a correct option. I got convinced while looking that region in the graph. But, the thing, I don't understand, why my initial approach had a loophole. The only possible explanation, that I can give you this, is : It's always true, that if $P\implies Q$ such that $Q$ is true, then, $P$ is true as well. But, the problem in my 1st attempt was with the parameterization itself as this parameterization, is not a strong one, and incomplete without further conditions. What do you think? (Also, I have edited my post) – Arthur May 01 '23 at 13:32
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    $P\implies Q$ and $Q$ being true does NOT mean $P$ has to be true. For e.g. consider $P=$ "Tom is a cat" and $Q=$ "Tom is a mammal". Now $P\implies Q$ and $Q$ could be true, but Tom need not be a cat, could be a monkey. – Macavity May 01 '23 at 13:32
  • @Macavity That's indeed a good one. Thanks for resolving my problems! I got the correct answers now. I am satisfied as I could justify it with the graph alone. The parametric solution was unnecessary here. As far as this mcq is concerned, it's a nice little trick, to solve mcqs. – Arthur May 01 '23 at 13:35

1 Answers1

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This isn't an answer to the question which has already been solved with the comments from Macavity, Gerry Myerson and linked documents. Below in addition and for information :

The equation $$x^y=y^x$$ can be analytically solved for $y(x)$ thanks to a special function.

$$e^{y\ln(x)}=e^{x\ln(y)}\quad\implies\quad y\ln(x)=x\ln(y)$$ $$-y\frac{\ln(x)}{x}=-\ln(y)$$ $$e^{-y\frac{\ln(x)}{x}}=\frac{1}{y}$$ $$y\:e^{-y\frac{\ln(x)}{x}}=1$$ $$-y\frac{\ln(x)}{x}\:e^{-y\frac{\ln(x)}{x}}=-\frac{\ln(x)}{x}$$ From the definition of the LambertW function https://mathworld.wolfram.com/LambertW-Function.html which is the inverse function of $\quad Y\:e^Y=X$ : $$Y=W(X)$$ With $\quad Y=-y\frac{\ln(x)}{x}\quad$ and $\quad X=-\frac{\ln(x)}{x}\quad$ we get : $$-y\frac{\ln(x)}{x}=W\left(-\frac{\ln(x)}{x} \right)$$ The analytic solution is : $$\boxed{y(x)=-\frac{x}{\ln(x)}W\left(-\frac{\ln(x)}{x} \right)}$$

The graph of $y(x)$ is shown below. The LambertW function is multivalued, noted $W_0$ and $W_{-1}$ in the real range. Here the symbol $W$ means the concatenation of the two real branchs $W_0$ and $W_{-1}$ in order to simplify the writing of the whole function.

As a consequence the function $y(x)$ is multivalued accordingly to $W_0$ for $0<y<e$ and $W_{-1}$ for $y>e$. The asymptote for the lower branche $x\to\infty$ is $y=1.$ The vertical asymptote for the uper branche $y\to\infty$ is $x=1.$

The curve would be symetrical wrt the first diagonal if the scales on the axis were the same.

enter image description here

Of course the graph isn't a rigorous mathematical proof but it can help to get an overall view of the behaviour.

JJacquelin
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  • $W(-\ln x/x)=-\ln x$... – Bob Dobbs Apr 30 '23 at 10:40
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    @Bob Dobbs . Of course $W_0(-\ln(x)/x)=-\ln(x)$ for $0<x<e$ this is a linear part of the branch of the function $y(x)$. But $W_0(-\ln(x)/x)\neq-\ln(x)$ for $x>e$. Also $W_{-1}(-\ln(x)/x)=-\ln(x)$ for $x>e$. But $W_{-1}(-\ln(x)/x)\neq-\ln(x)$ for $1<x<e$. – JJacquelin Apr 30 '23 at 13:22
  • Then what is equal to otherwise? – Bob Dobbs Apr 30 '23 at 18:57
  • @Bob Dobbs : Sorry, I don't understand your question "Then what is equal to otherwise?" . – JJacquelin May 01 '23 at 08:35
  • @Jacquelin : Is there an expression for $W_0(-\ln(x)/x)$ when $x>e$? – Bob Dobbs May 01 '23 at 10:52
  • @JJacquelin Thank you for your answer! But unfortunately, I haven't learnt "Lambert functions." I would be grateful if you could kindly point out the portion where my reasonings in the original post goes faulty. – Arthur May 01 '23 at 11:32
  • @Fdst Zfsy ; The analytical solving involving the Lambert function isn't given as a direct answer to the original question. This is just out of curiosity and for specific information. Sorry, I have nothing more to add to the lot of comments that you already got to your question. – JJacquelin May 01 '23 at 12:40
  • @Bob Dobbs . The common use of a special function is to give a closed form to a function which cannot be expessed with the combination of a finite number of elementary functions. This is the case of the Lambert function. The closed form of $W_0(-\ln(x)/x)$ is the function $W_0(-\ln(x)/x)$ itself. Sometimes a special function can be reduced to elementary functions in a limited range of the argument but not in the whole range of the argument (If not, this would no longer be a special function). This is what happens in the present case. – JJacquelin May 01 '23 at 13:01
  • @JJacquelin I know. But maybe another form of expression. Not so closed. I just wonder about the hyperbolic shape of the second solution. I asked too much. Sorry. – Bob Dobbs May 01 '23 at 13:06
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    @Bob Dobbs . Probably some approximate solution might be found in form of series. This should be a new interesting question to raise on the forum. – JJacquelin May 01 '23 at 13:16
  • @JJacquelin Thanks! I edited my post. The issue is resolved! – Arthur May 01 '23 at 13:40
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    @Fdst Zfsy : All's well that ends well. – JJacquelin May 01 '23 at 13:44
  • @JJacquelin Indeed.Couldn't been written more correctly! – Arthur May 01 '23 at 13:46