probability theory - Find a monotone function $u(x)$ such that a random variable has a uniform distribution

Suppose $X$ has density (pdf) $f_X(x) = \frac{2}{3}(x-1)$ for $2 < x < 3$.

a.) Find the pdf of $Y = -\log X$. (Make sure to specify the range and validity of your answer).

b.) Find a montone function $u(x)$ such that the random variable $Z = u(x)$ has a Uniform$(0,1)$ distribution.

Solution a.) We have from Casella and Berger theorem 2.1.8 $$f_Y(y) = \sum_{i=1}^{n}f_X(g^{-1}_i(y))\left|\frac{d}{dy}g_i^{-1}(y)\right|$$
We are given $$Y = -\log X \Rightarrow g^{-1}(y) = e^{-y}$$
Thus

\begin{align*}
f_Y(y) = f_X(g^{-1}(y))\left|\frac{d}{dy}g^{-1}(y)\right|\\
&= \frac{2}{3}(e^{-y} - 1)e^{-y}\\
&= \frac{2}{3}e^{-y}(e^{-y} - 1) \ \ \ \ \ \ -\log(3) < y < -\log(2)
\end{align*}
To verify this is correct note that $$\int_{-\log(3)}^{-\log(2)} \frac{2}{3}e^{-y}(e^{-y} - 1) dy = 1$$

I am not sure how to do part b any suggestion is greatly appreciated.

Answer

See that $\Pr(Z\leq z)=z$ if $Z$ is uniformly distributed. Now see that:

$$
\Pr(Z\leq z)=\Pr(u(X)\leq z)=\Pr(X\leq u^{-1}(z))\\
=\int_2^{ u^{-1}(z)}\frac 23(x-1)dx=\frac{u^{-1}(z)^2}{3}-\frac{2u^{-1}(z)}{3}
$$
So we have:
$$
\frac{u^{-1}(z)^2}{3}-\frac{2u^{-1}(z)}{3}=z.
$$
which means that:
$$

u(x)=\frac{x^2}{3}-\frac{2x}{3}.
$$