Transformation of Uniform Random Variable Let $X\sim \mathcal{U}(0,1)$  and $Y = \ln\left((1-X)^{-c}\right)$ for some $c>0$.
We need to show that $Y \sim {\mathrm{Exp}}(1/c)$.
We got to here but couldn't go further:
$$
   \mathbb{P}(Y \leqslant y) = \mathbb{P}( X \leqslant 1- \mathrm{e}^{-c y})
$$
Thanks in advance.
 A: You're almost there. Since $X$ is uniform, $\mathbb{P}(X \leqslant x) = x$ for all $0\leqslant x \leqslant 1$. Thus you established that
$$
   F_Y(y) = \mathbb{P}(Y \leqslant y) = \mathbb{P}\left(X \leqslant 1-\mathrm{e}^{-\frac{y}{c}} \right) = 1 - \exp\left(- \frac{y}{c} \right)
$$
Now compare this with CDF of exponential distribution with parameter $\lambda$:
$$
    F_{\mathrm{Exp}(\lambda)}(y) = 1 - \exp(-\lambda y)
$$
This proves that $Y \sim \mathrm{Exp}(c^{-1})$.
A: $\mathbb{P}\left( {Y \leqslant y} \right) = \mathbb{P}\left( {\ln \frac{1}
{{{{\left( {1 - X} \right)}^c}}} \leqslant y} \right) = \mathbb{P}\left( {X \leqslant 1 - {e^{ - \frac{y}
{c}}}} \right) = \int_0^{1 - {e^{ - \frac{y}
{c}}}} {\frac{1}
{{1 - 0}}dx}  = 1 - {e^{ - \frac{y}
{c}}}$
Since CDF of exponentially distributed variable $Z$ with parameter $\lambda $ is given by $\mathbb{P}\left( {Z \leqslant z} \right) = 1 - {e^{ - \lambda z}}$, we conclude that  $Y$ has exponential distribution with parameter $\frac{1}{c}$
A: Recall that the derivative of   the distribution function of a continuous random variable $Y$ is the density of $Y$.
You've found that $F_Y(y)=1-e^{-y/c}$ for $y\ge0$.  
Also $f_Y(y)=0$ for $y<0$ (since  $ Y=\ln \bigl( \, (1-X)^c \bigr)\,)\ge 0 $ ). So
$$
F_Y(y)= \cases { 1-e^{-y/c}&, y\ge0\cr0&,\text{otherwise}  }
$$
Differentiating $F_Y(y)$ gives
$$
f(y) 
=\cases{{1\over c}e^{-y/c}&, y\ge 0\cr 0\vphantom{1\over c}&,\text{otherwise} . }
$$
So $f_Y$ is the density of an exponential variable with parameter $1/c$.
A: Here is a different approach. Given that $X\sim \mathcal{U}(0,1)$, $f_X(x)=1$.
From $y = \ln\left((1-x)^{-c}\right)$ we get $x=1-\mathrm{e}^{-\frac{y}{c}}$ and thus, $$\frac{\text{d}x}{\text{d}y}=\frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~~.$$ Thus 
$$\begin{align*}
f_Y(y) & = f_X\left(1-\mathrm{e}^{-\frac{y}{c}}\right)\left|\frac{\text{d}x}{\text{d}y}\right|\\
& = \frac{1}{c}\mathrm{e}^{-\frac{y}{c}}~~,
\end{align*} $$  which is what we wanted.
