What are the steps to isolate x in this equation?

y^x = x^y
xln(y) = yln(x)
x/ln(x) = y/ln(y)

Edit: whoops, I didn’t read thoroughly. You want only one x, not x’s and y’s separated. Sorry about that.

It’s not always possible to isolate a variable in an equation, so that could possibly be the case here.

@Mariah Can you give an example of when it is not possible to isolate a variable? I’m not saying that you’re wrong, but that I have never encountered that before.

At least according to the professor who taught my most recent calculus class, it is not possible to isolate y in this equation: e^(xy) + sqrt(x+y) – 4 = 0.

Your example seems like it should be possible, though. I’ll keep thinking about it.

The reason that you can’t do what you want is that your equation involves the transcendental functions exp and ln.

Alright, so maybe we can’t isolate x, but we can still pick apart the relationship between x and y. Obviously the equation is satisfied when x = y (substitute x in for y to get x^x = x^x). But since, as you remarked above, there are two possible x’s for y=4, there’s got to be at least one other relationship out there.

y^x = x^y
xln(y) = yln(x)
x/ln(x) = y/ln(y)
x/y = ln(x)/ln(y)

This shows that the equation is also satisfied when each variable equals its own natural log, i.e. x = ln(x), or e^x = x. Also note that it is arbitrary that I used the natural log, because any base log is legal to use here, so if we call the log base “b,” this equation is satisfied anytime b^x = x and b^y = y. I’m not sure how useful this is because I’m having trouble, just off the top of my head, even determining if b^x = x even has any real number solutions, but I figured I’d throw it out there in case it might give you any ideas as to where to go next.

@Mariah b^x = x has real solutions. You can change this to b = x^(1/x). If x=2, then b is the square root of two, and the equation works out. If x=3, then b is the cube root of three, etc.

@PhiNotPi Right, great!

Also, check this link, it answers your question and is very interesting.

I asked this question some time ago and got a great answer from BonusQuestion, but I do not remember it. You can do a search here to see if you can find it.

One way of attacking this is to set y=ax.
We get (ax)^x = x^(ax)
(ax)^x = (x^a)^x
So one solution would be ax = x^a, x^(a – 1) = a, x = (a – 1) root of a
For example, if a=2, then x=2 and we get x=2 and y=4
I have the feeling though that this method does not yield all the solutions.

@LostInParadise I wonder if that always gives the second, non-trivial answer.

If we set “a” to be some number, then x is given by
x = a^(1/(a-1)), as you have said. This means that
y = a * a^(1/(a-1))
y = a^(1 + 1/(a-1))
y = a^((a-1)/(a-1) + 1/(a-1))
y = a^(a/(a-1))

I don’t know if this means anything.

This formula always gives a nontrivial answer for A != 1. It is very clear why.

To prove that formula gives all non-trivial answers, we have to prove that there is only one answer with a given ratio between x and y.

Here is the Wolfram solution
And here is the product log function they talk about, which I had never heard of before.

How does Wolfram manage to do this? It is almost scary.

It seems that, for positive real numbers at least, a combination of the two methods we have found gives all of the answers.

For a given number “A”, the numbers a^(1/(a-1)) and a^(a/(a-1)) make a pair, and those numbers with themselves make a pair.

This only gives all of the positive solutions. For A=2, then the numbers (2,4), (2,2), and (4,4) are given as solutions. It does not give negative solutions, such as (-2,-4), where A=2.

The solutions (2,2) and (4,4) overlap the solutions for other values of A.
It is very likely that a^(1/(a-1)) = b^(b/(b-1)), for some other number b. This gives the same solution twice.

To eliminate this, we can throw away the a^(a/(a-1)) pair, which will be covered by another pair for a different A.

@LostInParadise Wolfram is indeed “scary-smart.” I never heard of ProductLog function, either. I like how the graph (in your 2nd link) of powers of i iteratively converges to the single value that solves i^z = z.

i have a headache.

@Dutchess_III You’re welcome.

I’ll see YOU in the chat room, mister!

@Dutchess_III Now I have a headache!

heh!

The closed form is

x = -y W(-ln(y) / y) / ln(y)

where W(x) is the Lambert W function. This function either has 0, 1, or 2 real solutions, depending on the value of -ln(y) / y and an infinite number of complex solutions.

For y=2, these include 4, 2, 9.09073–21.508 i, 10.7467–40.0345 i, etc.