Solving equation with exponentials

Question

How to solve $ {z = x^y, x = y^z, y = z^x }$ for $ x, y $ and $ z? $

Is some sort of triple Lambert W to be introduced?

Done so far:

Taking logs,

$$ \log z = y \log x , \log x = z \log y , \log y = x \log z ; $$

Plug the third into second and that again into the first and simplify to get cyclically:

$$ x\, y \, z = x\, y \cdot log _y x = y\, z\cdot log _z y = z\, x\cdot log _x z = 1 ; $$

Then, how to proceed next?

Plots of $y= x^x$ and and $x \cdot y\, log_y x = 1 $ pass through (0,1) and (1,1).

So, is (x,y,z) = (1,1,1) the only solution?

My motivation for the post:

Solution of $ x^2 = 2^x $ has one more real root apart from inspection originated

solution x=2. Similarly there could be some other, may be complex roots at least.

Any solution of $a = a^a$ would do, though of course finding complex solutions for the infinite tetration doesn't look easy. — Macavity, Jan 15 '15 at 17:41

score 0 · Answer 1 · answered Jan 15 '15 at 21:06

0

Assuming all three variables are positive reals, let us rewrite the three equations as $\quad\sqrt[{\LARGE z}]z=y^y,$ $\sqrt[{\LARGE y}]y=x^x,\quad\sqrt[{\LARGE x}]x=z^z.~$ Then let us extract one expression for $z=z_1(x)$ from the last equation, and another one from the first two equations, $z=z(y)=z_2(x)$, since $y=y(x)$ from the second equation. Plotting them, we see that the two graphics intersect only for $x=1$.

answered Jan 15 '15 at 21:06

Lucian

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In fact, the first diagonal is tangent to both graphics at $x=1$, each graphic being completely contained within each of the two half-planes determined by the diagonal in question. – Lucian Jan 15 '15 at 21:51

Solving equation with exponentials

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