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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!

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

1 Answer 1

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There is indeed no such function

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.

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  • $\begingroup$ Thank you so much ! $\endgroup$
    – Last X
    Feb 20 at 13:53

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