The rectangle $ABCD$ with $AB>BC$ is cut into six smaller rectangles, each of which is similar to $ABCD$. The figure is drawn to scale, and each number is the area of the rectangle it labels. The areas are $32,\,16,\,8,\,4,\,2,\,2$. What is the length $AB$?

1. Let $k>1$ be the ratio of the long side to the short side of $ABCD$; every rectangle in the figure has this ratio. Let $BC=h$, so $AB=kh$ and $[ABCD]=kh^2$.
2. The rectangle labeled $32$ stands on its short side and has the full height $h$, so its width is $h/k$ and its area is $h\cdot\dfrac hk=\dfrac{h^2}k$.
3. The six areas add up to $32+16+8+4+2+2=64$, so $kh^2=64$. Comparing with $\dfrac{h^2}{k}=32$ gives $k^2=\dfrac{kh^2}{h^2/k}=2$, i.e. $k=\sqrt2$.
4. Then $h^2=\dfrac{64}{\sqrt2}$ and $AB^2=k^2h^2=2\cdot\dfrac{64}{\sqrt2}=64\sqrt2$, so $AB^4=64^2\cdot2=2^{13}$.
5. Hence $AB=2^{13/4}=\boxed{8\sqrt[4]{2}}$.