Vietnam National Olympiad Problem 1

Since $x^2+y^3=29\quad \Rightarrow y = \sqrt [3] {29-x^2}$ .

$\Rightarrow \log_3 {x} \log_2 {y} = 1 \quad \Rightarrow \dfrac {\log {x} \log {y}}{ \log {3} \log {2}} = 1 \\ \Rightarrow \log {x} \log {\sqrt [3] {29-x^2}} = \log {3} \log {2}$

Plotting $f(x) = \log {x} \log {\sqrt [3] {29-x^2}} - \log {3} \log {2}$ , we get the following graph and see that there are $\boxed{2}$ roots $x \approx 2$ and $x \approx 5$ .

excellent problem