# Find the number of solutions for $\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$ with a given binary operation.

On the interval $$(-1, 1)$$, consider the binary operation

$$x*y=\dfrac{2xy+3(x+y)+2}{3xy+2(x+y)+3}$$

with $$x, y \in (-1, 1)$$. I have to find the number of solutions for the equation:

$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$

Finding the left-hand side would be incredibly painful, so I didn't try that. I looked at the previous sub-point of this exercise and it looked like it might help, but I don't know how to use it exactly. In this previous sub-point I showed that for the function:

$$f:(-1, 1) \rightarrow (0, \infty) \hspace{2cm} f(x) = \dfrac{1}{5} \cdot \dfrac{1-x}{1+x}$$

it is true that:

$$f(x * y) = f(x)f(y)$$

$$\forall x,y \in (-1, 1)$$.

Still I don't know how to find the number of solutions for:

$$\underbrace{x*x*\ldots*x}_{x\text{ 10 times}}=\frac{1}{10}$$

• Hint: Apply $f$ both sides of the equation. – Peter Foreman Mar 18 at 19:18
• Is the star operation associative ? This could be checked in about 10 minutes of suffering. If so, then it would combine well with Peter Foreman's hint. – Simon Mar 18 at 19:41
• Actually, you don't need general associativity, you only need it in the case where every operand is equal, i.e. to $x$. This "restricted associativity" follows from commutativity of the star operation, which can be seen immediately to hold. – Simon Mar 18 at 20:03
• You have associativity in general. $f((x*y)*z) = f(x*y)f(z) = (f(x)f(y))f(z) = f(x)(f(y)f(z)) = f(x)f(y*z) = f(x*(y*z))$ and $f$ is a bijection, so $(x*y)*z = x*(y*z)$. – Jair Taylor Mar 18 at 20:54
• This is an example of a formal group law – Jair Taylor Mar 18 at 20:56

What a strange operation! The trick is to find an isomorphism, and they have given you the hint: because $$f$$ is a bijection, we can say that the operation of $$*$$ on $$(-1, 1)$$ is isomorphic to the operation of usual multiplication on $$(0, \infty)$$. Thus, in terms of $$y = f(x) \in (0, +\infty)$$, the problem is simply to solve $$y^{10} = f(1/10) \in (0, +\infty)$$, which has exactly one solution.
(One should check, quite painlessly, that having an isomorphism is "really this good"; for example, that we also get $$x * ... * x = f^{-1}(f(x)^n)$$ and so on. Indeed, a bijection satisfying the homomorphism law is the correct notion of an isomorphism between magmas.)
• @CristopherGadzinski How did you find that $y^{10} = f(1/10)$ has only one solution? – user1502 Mar 19 at 21:44