If r>0 and (9r/2)^(1/3) = (1/2) r, what is the value of r?

Two good ways to go here. First, the algebra:

    \begin{align*}\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}&=\dfrac{1}{2}r\\\\\left(\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}\right)^3&=\left(\dfrac{1}{2}r\right)^3\\\\\dfrac{9r}{2}&=\dfrac{1}{8}r^3\\\\36r&=r^3\end{align*}

One possible solution to the above is r = 0, but the question says that r > 0, so we can ignore that and move on:

    \begin{align*}36r&=r^3\\36&=r^2\\6&=r\end{align*}

We need not consider r = –6 because we know r is positive.

The other way to go is to graph both sides of the equation and find the intersection(s).

Because r is positive (my software made me use x in the graph above, as your calculator probably will, so just remember that xr), the only intersection that matters is (6, 3). At that intersection, x = 6, so we know that the value of r is 6.