How can one solve this equation?

Question

$$c = x \left( 1 - \left( 1- \frac{1}{x} \right)^a \right)$$

where $c$ and $a$ are constants. How can one solve this equation?

Answer 1

Obviously $x\ne 0.$ If $a$ is not an integer, then we should assume that $1-\frac{1}{x}>0$, so $\frac{1}{x}<1$ and $x\in(-\infty,0)\cup(1,\infty).$ Uur equation reads as $$\frac{c}{x}=1-\left(1-\frac{1}{x}\right)^a.$$ Next, if $c=0$, then $$1-\frac{1}{x}=1,$$ which is impossible. After next rearrangement we arrive at $$\frac{1}{c}\left[1-\left(1-\frac{1}{x}\right)^a\right]=\frac{1}{x}.$$ Put $C=\frac{1}{c}$ and $t=\frac{1}{x}.$ Then $t$ is a fixed point of $f(t)=C(1-(1-t)^a).$

Answer 2

Note that

$$c = x \left( 1 - \left( 1- \frac{1}{x} \right)^a \right)\iff \frac c x=1 - \left( 1- \frac{1}{x} \right)^a\iff1- \frac{c}{x}=\left( 1- \frac{1}{x} \right)^a\\\iff c\left( 1- \frac{1}{x} \right)-c+1=\left( 1- \frac{1}{x} \right)^a\iff\frac{\left( 1- \frac{1}{x} \right)^a-1}{\left( 1- \frac{1}{x} \right)-1}=c$$

and by $y= 1- \frac{1}{x}$

$$f(y)=\frac{y^a-1}{y-1}=c$$

Answer 3

Put $$1-\dfrac{1}{x}=t\iff x=\frac{1}{1-t}\Rightarrow c(1-t)=1-t^a$$ Therefore your equation is equivalent to a trinomial $$t^a-ct+(c-1)=0$$ from which the difficulty is clear for arbitrary constant $a$. You do have to apply numerical techniques when $a\notin\{0,1,2\}$ .