 Solution

For rate $\lambda >0$, returnFor $x\ge 0$, $\mathbb{P}(X>x)=\mathbb{P}(U_1<e^{-\lambda x})=e^{-\lambda x}$, so $X$ has the exponential distribution of rate $\lambda$. This is inverse transform sampling; using $-\log (1-U_1)/\lambda$ is equivalent. Uniform draws at exact endpoints can be excluded, since those events have probability zero under the ideal continuous law.