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?
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?
