# Solving y=x^x for x

Problem:
I'm trying to figure out how to solve $$y=x^x$$ for x as a function of y.

Related Equations:
$$\log_n a^b=b \ast \log_n a$$

Attempt at Solution:
I took the natural logarithm of both sides and got: $$\ln y=x \ast \ln x$$ I don't really have any idea where to go from here.

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Dr Transport
Gold Member

Remember that the $$\ln(x)$$ is nonsense for $$x \leq 0$$.

Remember that the $$\ln(x)$$ is nonsense for $$x \leq 0$$.
Uh... by plot, do you mean graph? I'm not trying to find where the graphs intersect or anything... I'm just trying to change the equation from $$y=f(x)$$ to $$x=f(y)$$

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Dr Transport
Gold Member
You can't......

You mean to say that $$x^x$$ has no inverse that can be defined in terms of elementary functions? Then what about using calculus? Can it be defined then?

HallsofIvy
Homework Helper
Depends upon what you mean by "using Calculus". The inverse of y= xx is the "Lambert W function".

By "using calculus," I mean, for example, defining the function with sigma notation, the derivative of something, etc. Here's a good example: the indefinite integral $$\int x^x dx$$ can not be defined in terms of elementary functions, but can be defined as the indefinite integral of $$x^x$$ So, what I'm asking is, can the inverse of $$x^x$$ be defined in terms of elementary functions as well as indefinite integrals, series, sums (i.e. sigma notation), derivatives, and so on?

I did some research on the "Lambert W function," and it's not the inverse of $$x^x$$. It's the inverse of $$x \ast e^x$$.

Nchimy

Solving y=x^x for x does have a solution in terms of Lambert W Function. Although this reply may have come late, one could still make use of it later.

To make the problem easy, lets assume that x and y are real. Then we can proceed as follows:

If y = x^x, .........(1)

then since we also have

x = exp[ln[x]], .........(2)

we may conviniently express (1) as

y=x^exp[ln[x]]. .........(3)

Now, taking logs on both sides of (3) (and noting that ln[d^c]=cln[d]) gives

ln[y] = ln[x] * exp[ln[x]] ..... (4).

Imediately, we see that (4) can be solved for ln[x], using Lambert W Function, as

ln[x] = W[ln[y]] ..... (5)

so that

x = exp[W[ln[y]]] .... (6) .

The solution in (6) is valid for y > 0 because ln[y] for real values of y makes sense only within this range. From (6) (and taking note that W=1while W[e] =1) we see that

(a) x = 0 when y = 1,
(b) x = 1 when y = e
(c) x is only real when ln[y] >= -(1/e) (or equivalently x is real for y >= exp[-(1/e)) but complex and multivalued otherwise.
(d) x increases monotonically with increasing value of y.
(e) The exists a taylor series expansion of x, about ln[y]=0, with a radius of converges equal to -(1/e).

For more details on th Lambert W function, please refer to Corless R M et al, “On the Lambert W function”, Adv. Comput. Math, Vol. 5, pp.329-359.

Cheers.....

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