How can I solve $x^x = 5$ for $x$?

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Is $x^x=y$ solvable for $x$?

I've been playing with this equation for a while now and can't figure out how to isolate $x$.

I've gotten to $x \ln x = \ln 5$, which seems like it would be easier to work with, but I can't figure out where to go from there.

Is it possible to solve this algebraically? If not, how can I find the value of $x$?

@J.M. Definitely. I figured it was asked before, but couldn't fathom how to search for it. — StrixVaria, Apr 17 '12 at 18:19
In adition to the answers: a quick numeric solution can be obtained iteratively via $x=5^{1/x}$. Or, more quick: $x= \sqrt{x , 5^{1/x}}$ — leonbloy, Apr 17 '12 at 19:15

score 5 · Accepted Answer · edited Dec 18 '12 at 10:28

We can find the result using the Lambert W function.

Let's define $y\,e^y=t$. Then $y=W(t)$ where $W(t)$ is the Lambert W function.

$$x=e^y\Rightarrow x^x= (e^y)^{e^y}=e^{y\,e^y}=5$$

$$\log e^{y\,e^y}=\log 5$$

$$y\,e^y\log e=y\,e^y=\log 5$$

Thus, $t=\log 5$, and from my first definition $y=W(\log 5)$, so $x=e^y=e^{W(\log 5)}$.

It can be expressed in another way too.

$$xy=y\,e^y=t=\log 5$$

$$x=\frac{\log 5}{y}=\frac{\log 5}{W(\log 5)}$$

I asked Wolfram Alpha what the numerical value is, and it said $x\approx2.129372$.

score 1 · Answer 2 · answered Apr 17 '12 at 18:11

1

It's not possible to solve this algebraically.

Look at Lambert W function. In your case, the solution is $x=\frac{\ln(5)}{W(\ln(5))}$.

answered Apr 17 '12 at 18:11

M.B.M.

2 Answers2