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My first approach involved simply taking the product: $$xy=a^bb^a=b^{b\log_b(a)+a}$$ However, I can't really do anything to, for instance, separate the exponent from the base $b$. Additionally, division has basically the same problem, and addition and subtraction are quite messy. Therefore, I settled on exponentiation and logarithms for my second and third approaches. $$x^y=({a^b})^{b^a}=a^{b^{a+1}}$$ $$\log_x(y)=\frac{\ln(y)}{\ln(x)}=\frac{a\ln(b)}{b\ln(a)}=\frac{a}{b}\log_a(b)$$ The third approach looked especially promising in conjunction with the first approach, but there was no way to surpass the $b^\mathrm{<exponent>}$ barrier.

EDIT

I am looking for algebraic solution, that can apply to real numbers, instead of number theory solutions, that can only apply to the integers. Thank you.

Zuter_242
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  • We limit ourselves to positive values. Note that $$a^b=x\text { has infinitely many solutions} (a,b)=(\sqrt[n] x,n)$$ and so is for $b^a=y$ whose solutions are $(b,a)=(\sqrt[n] y,n)$. Can you finish now? – Piquito Jun 03 '22 at 18:10
  • @Piquito How does your approach solve the problem? – M. Wind Jun 04 '22 at 16:37
  • @M. Wind: Put in the second set of given solutions $m$ instead of $n$ (they are arbitrary) and try to go ahead. – Piquito Jun 04 '22 at 19:59

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