Let f(x)
Question:

Let $f(x)=\frac{x}{x-1}$ and $\frac{f(\alpha)}{f(\alpha+1)}=f\left(\alpha^{k}\right)$, then $k=$  ______________.

Solution:

Given: $f(x)=\frac{x}{x-1}$ and $\frac{f(\alpha)}{f(\alpha+1)}=f\left(\alpha^{k}\right)$

$\frac{f(a)}{f(a+1)}=\frac{\frac{a}{a-1}}{\frac{a+1}{a+1-1}}$

$=\frac{\frac{a}{a-1}}{\frac{a+1}{a}}$

$=\frac{a^{2}}{(a-1)(a+1)}$

$=\frac{a^{2}}{a^{2}-1}$        …(1)

It is given that,

$\frac{f(a)}{f(a+1)}=f\left(a^{k}\right)$

$\Rightarrow \frac{a^{2}}{a^{2}-1}=\frac{a^{k}}{a^{k}-1}$

$\Rightarrow k=2$

Hence, k = 2.