Introduction to statistics Let $X$ be uniformly distributed on $[a,b]$. Find a function $f$ such that $Y=f(X)$ is exponentially distributed with parameter $1$. How do I do this?

You need $\Pr(f(X)\le y) = 1-e^{-y}$ for $y>0$.  That is true if $f$ is an increasing function and $\Pr(X\le f^{-1}(y))=1-e^{-y}$.  But you know that $\Pr(X\le x)=(x-a)/(b-a)$ if $a\le x\le b$, so $\Pr(X\le f^{-1}(y))=(f^{-1}(y)-a)/(b-a)$.  Thus it is enough to find a fuunction $f$ for which
$$
\frac{f^{-1}(y)-a}{b-a} = 1-e^{-y}\text{ for }y\ge 0. 
$$
Can you take it from there?
