Don't do this by hit and trial!!

Taking the natural log of both sides and rearranging, we find that

$\dfrac{\ln(x)}{x} = \dfrac{\ln(y)}{y}$ .

So now look at the function $f(x) = \dfrac{\ln(x)}{x}$ for $x \ge 1$ . We have $f(1) = 0$ and $f'(x) = \frac{1 - \ln(x)}{x^{2}} = 0$ when $x = e = 2.71828....$ . So $f(x)$ is increasing on the interval $[1,e]$ , where it peaks at a value of $f(e) = \frac{1}{e}$ and then decreases gradually to $0$ as $x \rightarrow \infty$ .

Given this shape of curve, we can only have $f(x) = f(y)$ with $1 \le x \lt y$ if $x$ lies on the interval $[1,e)$ and $y$ lies on the interval $(e, \infty)$ . So if $x$ is to be a positive integer $\ne 1$ then it can only be $2$ .

Now by observation we have $2^{4} = 4^{2}$ , and by the above discussion we can conclude that this is the only integer solution. However, since we are asked for the number of ordered pairs we can have both the ordered pairs $(2,4)$ and $(4,2)$ as solutions, giving us a total of $\boxed{2}$ solutions.

Consider $y>x$ , thus notice that x divides y . Now let $y=ax$ for some positive integers of $a>1$ , hence $x^{ax}=(ax)^{x}$ . Expanding the equation yields $a=x^{a-1}$ , thus $a>x$ or $a=x$ . If $a=x$ , by comparison $1=a-1$ . Therefore, $a=x=2$ and this gives $y=4$ . If $a>x$ , $a-1>1$ or $a>2$ yields x=a^{ $\frac{1}{a-1}$ } which does not give positive integers of x for $a>2$ . Now consider $x>y$ , we shall get $x=4$ and $y=2$ , thus the number of ordered pairs of distinct positive integers $(x,y)$ is $\boxed{2}$ .

there are 2 pairs,(2,4) and (4,2) substitute in the equation 16=16.answer is therefore 2 pairs.

only 0 and one satisfy the expression.

What a simple question this is!