From the given equation, \[\frac{\tan (x + 100^\circ)}{\tan (x - 50^\circ)} = \tan (x + 50^\circ) \tan x.\]Then \[\frac{\sin (x + 100^\circ) \cos (x - 50^\circ)}{\cos (x + 100^\circ) \sin (x - 50^\circ)} = \frac{\sin (x + 50^\circ) \sin x}{\cos (x + 50^\circ) \cos x}.\]By Componendo and Dividendo, \[\frac{\sin (x + 100^\circ) \cos (x - 50^\circ) + \cos (x + 100^\circ) \sin (x - 50^\circ)}{\sin (x + 100^\circ) \cos (x - 50^\circ) - \cos (x + 100^\circ) \sin (x - 50^\circ)} = \frac{\sin (x + 50^\circ) \sin x + \cos (x + 50^\circ) \cos x}{\sin (x + 50^\circ) \sin x - \cos (x + 50^\circ) \cos x}.\]Applying the sum-to-product formula, we get \[\frac{\sin (2x + 50^\circ)}{\sin 150^\circ} = \frac{\cos 50^\circ}{-\cos (2x + 50^\circ)}.\]Hence, \[-\sin (2x + 50^\circ) \cos (2x + 50^\circ) = \cos 50^\circ \sin 150^\circ = \frac{1}{2} \cos 50^\circ.\]Then \[-2 \sin (2x + 50^\circ) \cos (2x + 50^\circ) = \cos 50^\circ.\]From double angle formula, we get $ \sin (4x + 100^\circ) = -\cos 50^\circ $. Since $ \sin (\theta + 90^\circ) = \cos \theta, $ \[\cos (4x + 10^\circ) = -\cos 50^\circ = \cos 130^\circ.\]This means $ 4x + 10^\circ $ and $ 130^\circ $ either add up to a multiple of $ 360^\circ, $ or differ by a multiple of $ 360^\circ $. Checking these cases, we find that the smallest positive angle $ x $ is $ \boxed{30^\circ} $.