Question

How many triplets $(x,y,z)$ of positive real numbers can be found such that $x^y=z,$ $y^z=x$ and $z^x=y$?

• A. 2
• B. 1
• C. 4
• D. 3

Answer 1

The correct option is B 1

$\Rightarrow$ $x^{xyz}=(x^y{)}^{xz}=(z^x{)}^z=y^z=x$

Thus, taking the base x logarithm, $xyz=1$.

Now, suppose $x\ge 1$, then $z=x^y\ge 1$ and $y=z^x\ge 1$.

Likewise, from $x\le 1$ follows that $y$, $z\le 1$.

That is, either $x,y,z$ are all $\ge 1$, or they are all $\le 1$.

Since their product has to be $1$, they are forced to be all equal to $1$, which gives a contradiction since they must be distinct.

Also, the above shows that, without the constraint that $x,y,z$ be distinct, the only solution is the triplet $(1,1,1).$