# check if there are such functions that verify the following functional equation

The statement of the problem: Determine if there are any functions $$f : (1, \infty)\rightarrow (1, \infty)$$ with the property:

$$x^{f(y)^x}$$ = $$y^{f(x)^y}\text{for every }x, y > 1$$

My approach: I first proved that it is an injective function. Then I logarithmized in the $$x$$ base and then in the $$y$$ base and I reached some equalities. From there, I tried to "guess" if there is any such function but I didn't find anything so I guess we have to prove that there is no such function.

Any and all proofs will be helpful. Thanks a lot!

• Mathjax tutorial: math.meta.stackexchange.com/questions/5020/… Feb 20 at 11:43
• The power scheme is ambiguous. Please use Mathjax. Feb 20 at 11:46
• it's x to the power of f(y) , and f(y) to the power of x . the same for the right side , it's y to the power of f(x) , and f(x) to the power of y . Feb 20 at 11:48
• After taking log on both sides substituting $y$ as $e$ , gives $f(x)$ right ?
– AAM
Feb 20 at 13:14
• @AAM there is $f(e)$ also, but it is just a constant
– D S
Feb 20 at 13:19

Suppose such $$f : (1, \infty)\rightarrow(1, \infty)$$ exists. Observe firstly that the dimensionality of the constraint on $$f$$ causes $$f$$ to be determined by its value at any one point in the following way. Let's consider, arbitrarily, $$f(2)$$. Let $$f(2) = a$$, then we have by the constraint for $$y = 2$$:
$$x^{(a^x)} = 2^{(f(x)^2)}$$ $$a^x \log_2 x = f(x)^2$$ $$f(x) = \sqrt{a^x \log_2 x}$$
Now we see if this holds up to scrutiny elsewhere along the constraint. Note the constraint is trivial if $$x = y$$, and of course it will hold for $$y = 2$$ (and by symmetry, $$x = 2$$) by design, so let's try, say $$x = 4$$ and $$y = 8$$. The constraint is then: $$4^{\left(\sqrt{a^8 \log_2 8}\right)^4} = 8^{\left(\sqrt{a^4 \log_2 4}\right)^8}$$ $$4^{\left(\sqrt{3a^8}\right)^4} = 8^{\left(\sqrt{2a^4}\right)^8}$$ $$4^{(9a^{16})} = 8^{(16a^{16})}$$ $$9a^{16}\log_2 4 = 16a^{16}\log_2 8$$ $$18a^{16} = 48a^{16}$$
Clearly this has the sole solution $$a = 0$$. Yet, by supposition, $$a$$ is in the range of $$f$$, but this contradicts $$f$$ being into $$(1, \infty)$$, thus no such $$f$$ exists.