What is the measure, in degrees, of the smallest positive angle $x$ for which $4^{\sin ^{2} x} \cdot 2^{\cos ^{2} x}=2 \sqrt[4]{8}$ ?
Using exponent and trigonometric laws,

$$
\begin{aligned}
4^{\sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2 \sqrt[4]{8} \\
\left(2^{2}\right)^{\sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2 \sqrt[4]{2^{3}} \\
2^{2 \sin ^{2} x} \cdot 2^{\cos ^{2} x} & =2^{1} \cdot 2^{3 / 4} \\
2^{2 \sin ^{2} x+\cos ^{2} x} & =2^{7 / 4} \\
2 \sin ^{2} x+\cos ^{2} x & =\frac{7}{4} \\
\sin ^{2} x+\left(\sin ^{2} x+\cos ^{2} x\right) & =\frac{7}{4} \\
\sin ^{2} x+1 & =\frac{7}{4} \\
\sin ^{2} x & =\frac{3}{4}
\end{aligned}
$$

Thus, $\sin x= \pm \frac{\sqrt{3}}{2}$.

The smallest positive angle for which one of these is true is $x=60^{\circ}$ (or $x=\frac{1}{3} \pi$ in radians).