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We are told that $f: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$

I know one method is to prove that a inverse exists, but I'm not 100% sure how to do that in this case. so instead I decided to create two functions $f(a,b)$ and $f(x,y)$ and set them equal to each other. $$f(a,b)=f(x,y)$$ $$12^a \cdot 18^b = 12^x \cdot 18^y$$ How exactly would be the next step to get $a,b = x,y$?

Also, if you could explain how to take the inverse of the function that would be much appreciated.

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    $\begingroup$ Hint: The order of $2$ on the left is $2a+b$, the order of $3$ is $a+2b$ $\endgroup$ – lulu Apr 4 at 23:21
  • $\begingroup$ Context clues suggest your function $f$ is a binary operation on the naturals $\mathbb{N}\times\mathbb{N}\to\mathbb{N}$, but it's important to specify the domain and codomain explicitly, since the extension of $f$ to the reals is not injective. $\endgroup$ – K B Dave Apr 4 at 23:39
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$f(a,b)=(2^2\times 3)^a(2\times 3^2)^b=2^{2a+b}\times 3^{a+2b}$.

$f(a,b)=f(x,y)$ is equivalent to $2a+b=2x+y$ and $3a+2b=3x+2y$

We deduce that $4a+2b=4x+2y$

$3a+2b=3x+2y$ You substract this two last equations you deduce that $a=x$ and $b=y$.

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