The random variable $log(\frac{X}{x_0})$ has an exponential distribution with parameter $\alpha$ It is said that a random variable $X$ has a Pareto distribution with parameters $x_0$ and $\alpha$ for $(x_0 > 0)$ and $(\alpha > 0)$ if $X$ has a continuous distribution for which the p.d.f. $f(x|x_0, \alpha)$ is as follows:
$$f(x|x_0, \alpha) = \frac{\alpha x_0^\alpha}{x^{\alpha +1}},\quad x\ge x_0, x_0>0, \alpha>0.$$
Show that if $X$ has this Pareto distribution, the random variable $\log(X/x_0)$ has an exponential distribution with parameter $\alpha$.
I took the approach of $Y = \log(X/x_0)$ and re-arranged for $X$ to get $X = x_0\exp(Y)$.
though I'm unsure of how to proceed from here. Furthermore, what actually explains that a relations exists between the Pareto distribution and an exponential distribution; for example, why is $\log(X/x_0)$ mentioned?
My further approach:
$x_0\exp(y)dy = dx$
then $$g(y) = f(x_0\exp(y)|x_0, \alpha) = \frac{\alpha x_0^{\alpha}}{(x_0\exp(y))^{\alpha+1}}x_0\exp(y)=\frac{\alpha}{\exp(y)^{\alpha}}=\alpha\exp(y)^{-\alpha}$$

Let $X\sim Pareto(x_0,\alpha)$, then the PDF is as you wrote and the CDF is
$$P(X\le k)=1-\left(\frac{x_0}{k}\right)^{\alpha}.$$
Now denote $Y=log\left(\frac{X}{x_0}\right)$, it only takes 5 transitions from $P(Y\le y)$ in order to show $Y\sim exp(\alpha)$:
$$P(Y\le y)=P\left( log\left(\frac{X}{x_0}\right) \le y \right)=P\left( \frac{X}{x_0}\le e^y \right)=P(X \le x_0e^y) \\= 1-\left(\frac{x_0}{x_0e^y}\right)^{\alpha}=1-e^{-\alpha y}$$
