Probability distribution of $e^X$, when $X ~$ uniform$(0,1)$ What is the probability distribution of $e^X$, when $X$ is a random variable and follows the uniform distribution $U(0,1)$?
I noticed that the probability of $e^X$ decays exponentially from $1$ to $e$, even though probability distribution of $U(0,1)$ is constant. Is there an explanation why this is so? Clearly, each number is as likely to be picked, and thus shouldn't each result be equally likely?

$$\because f(x) =
\begin{cases}
1,  & 0\leqslant x \leqslant 1 \\[2ex]
0, & others
\end{cases}
$$
And $Y = e^{X}$
$F_{Y}(x)=P(Y\leqslant x) = P(e^{X}\leqslant x) =P(X\leqslant ln(x))$;
When $ x \leqslant1, F_{Y}(x) = 0$ , $\therefore f_{Y}(x) = 0$.
When $  1\lt x \lt e, F_{Y}(x) = \int_{-\infty}^{lnx} 1* dt = \int_0^{lnx} 1* dt = lnx ,  \therefore f_{Y}(x) = 1/x $
When $x > e, F_{Y}(x) = 1,  \therefore f_{Y}(x) = 0$
$$\therefore f_{Y}(x) =
\begin{cases}
\frac{1}{x},  & 0\lt x \lt e \\[2ex]
0, & others
\end{cases}
$$
