How can we find x for x^n = n^x

Question

Find values of x such that $x^n=n^x$ Here, n $\in$ I.

One solution will remain x=n But i want to find if any more solutions can exist

$$x^n=n^x$$

Answer 1

I love this question! I first saw it while in sixth form (I'm from London that means when I was 18)

The first thing we can do is to try and get just x on one side and just y on the other side. Heres what we can do using logs:
If x^y = y^x then we have to have log(x^y ) = log(y^x) as the log function is a bijection.
I.e: x^y = y^x $\iff$ log(x^y ) = log(y^x).

(Please note that when I write log I mean the natural based log, you may be used to seeing ln instead but I mean the very same thing.)

Now we can use the following log rule : log(a^b )=b* log(a) to get
x^y = y^x $\iff$ y* log(x) = x*log(y)
Dividing both sides of our RHS by x *y yeilds:
x^y = y^x $\iff$ $\frac{log(x)}{x} = \frac{log(y)}{y} $
Perfect! We now have just x on one side and just y on the other. Lets call f(t) the function $\frac{log(t)}{t}$ so we know that the if any pair x and y solve x^y = y^x then we MUST also have f(x) = f(y) and likewise if we find any pair x and y such that f(x) = f(y) then we also know that x^y = y^x.

Onto finding soloutions of f(x) = f(y).
I will highly encourage you to graph this function either by hand or using something like Desmos and from that you'll be able to nearly get all thats left but heres a short summary of what you'll find:
1) The function tends to negative infinity as t tends to 0.
2) The function tends to 0 as t tends to infinity
3)There is a turning point (global maximum) of this function at t = e (this "little fact" is actually a great proof of why e^x > x^e for all x)
4) The function is strictly increasing up to e and then strictly decreasing after it.
1) and 2) can be seen from a simple look at limits and 3) and 4) come from a quick examination of its derivative.

So if we are after pairs of numbers x and y such that f(x) = f(y) we know that one of them needs to be less than e and one of them greater than e. If its only integer soloutions you are after, great! You only have 2 integers to check as the only ones less than e are 1 and 2. Its quite quick to see that 1 is not going to be helpfull at all as 1^x = 1 and x¹ = x so your only soloution if you chose x = 1 will be y =1 . Trying x =2 is more more fruitfull, a little checking and guessing gives x = 2 and y =4 a soloution to this, there are also no other y for x=2 as our f is strictly decreasing beyond e.
Thus the only integer soloutions to x^y = y^x is x =2 and y=4 (well you can have x =4 and y=2 but thats the same thing!)
If you are after non integer soloutions you have a few options, you can pick any x you want, as there will always be a y with the same value of f and then using some method of approximation the root to f(t)-f(x)

I hope this helped :)
Oskar

Answer 2

Welcome to the wonderful world of Lambert function !

If, on the search bar, you just type Lambert, you will find almost 3000 entries.

This function $W(z)$ cannot be expressed in terms of elementary functions but it is rather simple since it is "just" the solution of the equation $$z=W(z)\,e^{W(z)}$$

Because of lack of time, I let you the pleasure of discovering it in the linked page. In the real domain, $W(z)$ is defined as long as $z \geq -\frac 1e$ and it has two branches, the secondary one existing when $-\frac 1e \leq z \lt 0$.

To make the story short, for the equation $$x^n=n^x$$ beside the trivial $x=n$ potentially exist two solutions given by $$x_1=-\frac{n }{\log (n)}W_0\left(-\frac{\log (n)}{n}\right)\qquad \text{and} \qquad -\frac{n }{\log (n)}W_{-1}\left(-\frac{\log (n)}{n}\right)$$

Since you want $n$ to be a positive integer, I give you below a list of values of the $W_0(.)$ since the other one cannot exist for natural numbers. $$\left( \begin{array}{cc} n & x \\ 2 & 2.00000 \\ 3 & 2.47805 \\ 4 & 2.00000 \\ 5 & 1.76492 \\ 6 & 1.62424 \\ 7 & 1.53014 \\ 8 & 1.46250 \\ 9 & 1.41138 \\ 10 & 1.37129 \\ 20 & 1.19624 \\ 30 & 1.13767 \\ 40 & 1.10754 \\ 50 & 1.08893 \\ 60 & 1.07620 \\ 70 & 1.06690 \\ 80 & 1.05977 \\ 90 & 1.05412 \\ 100 & 1.04952 \end{array} \right)$$

To check it, use your pocket calculator for $n=50$ and $x=1.08893$

$$1.08893^{50}=70.7943\qquad \text{and} \qquad 50^{1.08893}=70.8043$$