What kind Logarithmic Base is this?

This equation can be written as

$(24\sin(x))^{\frac{3}{2}} = 24\cos(x) \Longrightarrow (24\sin(x))^{3} = (24\cos(x))^{2}$

$\Longrightarrow 24\sin^{3}(x) = \cos^{2}(x) = 1 - \sin^{2}(x)$

$\Longrightarrow 24\sin^{3}(x) + \sin^{2}(x) - 1 = 0$

$\Longrightarrow (3\sin(x) - 1)(8\sin^{2}(x) + 3\sin(x) + 1) = 0.$

Since there are no real roots for the quadratic, the only solution is $\sin(x) = \dfrac{1}{3}.$

This gives us that $\cos(x) = \pm \dfrac{2\sqrt{2}}{3}$ and thus $\cot^{2}(x) = 8.$

So the desired solution is $24*8 = \boxed{192}.$

Follow Brian's solution up to sin(x) = 1/3. Therefore, 24sin(x)^3 = 24cos(x)^2. Divide by 24sin(x)^2 to get 24sin(x) = cot(x)^2, which in turn leads to 24cot(x)^2 = 24x24x1/3 = 192.