5
$\begingroup$

I want to solve $$x^{x^3}=729$$

so I tried like below: $$\log_9{x^{x^3}}=\log_9{729}\\x^3\log_9x=3\\x^3=\frac{3}{\log_9x}\\x^3=3\log_x{9}\\\cdots$$

but I got stumped. Then I tried this: $$x^{x^3}=729\\(x^{x^3}=729)^3\\x^{3x^3}=(9^3)^3$$

It can be rewritten as: $$(x^3)^{x^3}=9^9\\x^3=9\\x=\sqrt[3]{9} $$

Then I checked it with Desmos: Plots of x^{x^3}-729 and x-9^(1/3)

It seems that I was correct. Anyway, I think my second try was somehow heuristic, and I'm asking for an analytic solution for that type of equation, if it exists. Or another point of view or idea.

$\endgroup$
3
  • 3
    $\begingroup$ I don't know what you mean by "analytic", but I think your second try was quite clever. If you change the problem just a little bit, say, to $x^{x^3}=728$, there might be no way to solve it, other than numerically (which is to say, approximately). $\endgroup$ Commented Nov 8 at 10:25
  • $\begingroup$ It’s not very hard just take log and use lambert w function $\endgroup$ Commented Nov 8 at 12:51
  • 1
    $\begingroup$ @AndreLin Or use base 9 maybe? $\endgroup$ Commented Nov 8 at 23:33

4 Answers 4

Reset to default
9
$\begingroup$

$$x^{(x^3)}=729=3^6$$ This leads to think that $x$ is on the form of $x=3^y$. If so $$(3^y)^{(3^{3y})}=3^6$$ $$3^{\left(y\:(3^{3y})\right)}=3^6\quad\implies\quad y\:(3^{3y})=6\quad\implies\quad (3y)\:(3^{3y})=18$$ Let $z=3y$ $$z\:3^z=18$$ This equation can be solved thanks to the LambertW function.

More simply we see that obviously $z=2$ is solution.

Then $y=\frac13 z=\frac23$ and finally with $x=3^y$ : $$x=3^{2/3}$$

$\endgroup$
0
6
$\begingroup$

More amusing would be $$x^{x^3}=k$$ Take logarithms $$x^3\log(x)=\log(k) \implies x^3\log(x^3)=3\log(k)$$ which gives $$x^3=\frac{3 \log (k)}{W(3 \log (k))}$$

Edit

Even more general, using the same gymnastic, $$\left(x^a\right)^{x^b}=k \quad \implies \quad x=\Bigg(\frac {\frac b a \log(k) } {W\left(\frac{b }{a}\log (k)\right) }\Bigg)^{\frac 1b}$$ where $(a,b)$ could be any real numbers

$\endgroup$
2
$\begingroup$

Your second method is perfectly valid, but you can solve it analytically using the Lambert W function. However, when using the Lambert W, you often have to use tricky manipulations similar to your second solution anyway, just to get the equation into a form that the Lambert W can handle.

$$\begin{align} x^{x^3} & = 729\\ x^{3(x^3)} & = 729^3 = 3^{18}\\ \end{align}$$

Let $y=x^3$ $$\begin{align} y^y & = 3^{18}\\ y\ln(y) & = 18\ln(3)\\ W(y\ln(y)) & = W(18\ln(3))\\ \ln(y) & = W(18\ln(3))\\ y & = 18\ln(3) / W(18\ln(3)) \end{align}$$

Of course, $y$ simplifies to $9$, thus $x = \sqrt[3]{9}$.

$\endgroup$
1
  • 1
    $\begingroup$ Here's a little calculator in Sage that can solve $x^{x^m}=z$ for arbitrary positive $m$ & $z$. sagecell.sagemath.org/?q=pytqyc $\endgroup$ Commented Nov 8 at 12:24
2
$\begingroup$

Render $729=9^3$. Then

$x^{3x^3}=y^y=9^9;y\equiv x^3$

So $y=9=3^2$ and $x=3^{2/3}$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.