= Solution
{c}
Let $r=1-p$. Since
$$
\frac{d}{dx}\log(1-r e^{-\beta x})
=\frac{\beta r e^{-\beta x}}{1-r e^{-\beta x}},
$$
integration from zero to $x>0$ gives $F(x)=(C/\beta)[\log(1-r e^{-\beta x})-\log p]$. Letting $x\to\infty$ and imposing unit total <probability> gives
$$
\boxed{C=-\frac{\beta}{\log p}>0,\qquad
F(x)=\begin{cases}
0,&x\leq0,\\
1-\dfrac{\log(1-(1-p)e^{-\beta x})}{\log p},&x>0.
\end{cases}}
$$
The <probability density function> is positive for $x>0$ because $0<(1-p)e^{-\beta x}<1$. Consequently the <cumulative distribution function> is increasing, starts at zero and tends to one. This verifies that the derived <normalizing constant> defines a <probability distribution>. The family is the <exponential-logarithmic distribution> with the complementary mixing parameter $r=1-p$.
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