smc

For these lengths to form a triangle of positive area, the Triangle Inequality tells us that we need $${\log }_{2}x+{\log }_{4}x>3$$ $${\log }_{2}x+3>{\log }_{4}x$$ $${\log }_{4}x+3>{\log }_{2}x.$$ The second inequality is redundant, as it's always less restrictive than the last inequality. Let's raise the first inequality to the power of $4$. This gives $4^{{\log }_{2}x}\cdot 4^{{\log }_{4}x}>64\Rightarrow {\left(2^2\right)}^{{\log }_{2}x}\cdot x>64\Rightarrow x^2\cdot x>64$. Thus, $x>4$. Doing the same for the third inequality gives $4^{{\log }_{4}x}\cdot 64>4^{{\log }_{2}x}\Rightarrow 64x>x^2\Rightarrow x<64$ (where we are allowed to divide both sides by $x$ since $x$ must be positive in order for the logarithms given in the problem statement to even have real values). Combining our results, $x$ is an integer strictly between $4$ and $64$, so the number of possible values of $x$ is $64-4-1=\boxed{59}$.