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