 This calculus video will be about the following problem. Can we plot all solutions of x to the y equals y to the x? So we start by thinking about trivial solutions. Well, there are some obvious solutions. We can just take x equals y. There's another solution, two to the power of four is equal to four to the power of two. So that gives us an extra two points underline. And in order to find all other solutions, I'm going to give two different methods. So the first method, let's just raise both sides to the power of one over xy. Then we find x to the one over x equals y to the one over y. So we've got to find values of x and y that have the same value of this function. Well, we can take the logarithms of both sides. So the logarithm of this is just the logarithm of x over x. And that has to be the logarithm of y over y. So whenever we've got powers, it's best to write them in terms of logarithms and exponentials. So obviously what we should do is we should draw the graph of log of x over x to see what it looks like. And the graph looks a bit like this. So it's equal to zero at x equals one at x less than one. It tends to minus infinity. And as x tends to infinity, it tends to zero because x is much bigger than log of x. So it looks something like this. Here's the point one. And obviously we should find out what the maxima of this function are. So we differentiate log of x over x and using the usual rule for derivative of a quotient. We find this is equal to one over x times x minus log of x times one or divided by x squared, which is equal to one minus log of x over x squared. So this vanishes when log of x is equal to one. In other words, when x is equal to e. So this is the point where x is equal to e. And you see this derivative is negative for x bigger than e. So it decreases all the way here and it's positive for x less than e. So the slope is always positive there. So that gives us a good idea of what this function looks like. Now we want to find points x and y that have the same value of this function. Well, obviously if x and y are the same, then they have the same value. So what about when x and y are not the same? Well, we could take x to be here and y to be up there. For example, and then if we make x a bit bigger, we find y is there and if x is a bit bigger, y is there and so on. So we can see that as x increases from one to infinity, the value of y decreases from infinity to one and they cross over at this point here where x and y are both e. So with this information, we can now plot the points where x to the y equals y to the x and the answer we get looks like this. We just take x and y axis and let's plot where x is equal to one and y is equal to one. And then first of all, we have the points where x equals y and then we have the points where x starts at one and increases to infinity and y starts at infinity and decreases down to one. So that ends up looking a bit like this. And as we saw, they meet at the point ee. So the points where x to the y equals y to the x consists of two components and they cross at this point here. So that gives one solution and I guess we can mark in the special values four, two and two, four. There's an alternative algebraic solution we can give. What we do is we make the substitution y equals t times x which is a very common and useful substitution. And then the equation x to the y equals y to the x becomes x to the tx equals tx to the x. Now we can take the x root of both sides and we find x to the t equals tx. Now we divide by x, x to the t minus one is equal to t. And now if t equals one, we just get the solution x equals y. If t is not equal to one, we can take the t minus one through to both sides. So we get x is equal to t to the one over t minus one. This is for t not equal to one. And y is equal to t times this which is equal to t to the t over t minus one. So as t varies from naught to infinity, this gives a parametric formula for this curve here. For example, if t is very large, then x is about one and y is about infinity. So here we have t tends to infinity and here we have t tends to zero. As t tends to one, the limit of t to the one over t minus one is e. So the limit is t tends to one of t to the one over t minus one is equal to e. So at t equals one, we get this point here. This point here is where t equals two, t equals a half, and we can do other things. For instance, if we take t equals three, we find x is equal to the square root of three, y is equal to three times the square root of three. So we get another point here, t equals three, point root three, three root three, and so on.