algebraic structure

[Vali]

On $\displaystyle (-1,1)$ I have an algebraic structure $\displaystyle x*y=\frac{2xy+3(x+y)+2}{3xy+2(x+y)+3}, x,y\in (-1,1)$

At first exercise I found the identity element ( -2/3).At second exercise I found that $\displaystyle f: (-1,1)\rightarrow (0,\infty ), f(x)=\frac{1-x}{5(1+x)}$ verifies $\displaystyle f(x*y)=f(x)\cdot f(y) \forall x,y\in (-1,1)$

I got stuck at the last exercise where I need to find the number of solutions of this equation: $\displaystyle x*x*x*...*x(ten \ times)=\frac{1}{10}$

I tried to use what I found at the second exercise: $\displaystyle f(x*x*x*...*x)=f(x)\cdot f(x)\cdot ...\cdot f(x)$ so
$\displaystyle f(x)^{10}=f(\frac{1}{10})$ so $\displaystyle f(x)=f(\frac{1}{10})^{\frac{1}{10}}$ so $\displaystyle x=f^{-1}(f(1/10)^{1/10})$
I found that $\displaystyle f^{-1}(x)=\frac{1-5x}{5x+1}$ and now I just find the value of $\displaystyle f(\frac{1}{10})^{\frac{1}{10}}$ and replace it in the inverse of the function ?I got some ugly calculations.Is my method correct?

 

[Steven G]

i would first compute x*x which will be the division of two quadratics. Then x*x*...*x = (x*x)⁵=1/10
So x*x= 1/10^(1/5) and this is easily solved.