1. **State the problem:** Solve the equation $\log_2 x + \log_4 x + \log_8 x = \frac{22}{3}$.\n\n2. **Recall the change of base formula:** For any logarithm, $\log_a b = \frac{\log_c b}{\log_c a}$. Here, we will express all logarithms in base 2 for consistency.\n\n3. **Rewrite each term:**\n- $\log_2 x$ remains as is.\n- $\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}$ since $\log_2 4 = 2$.\n- $\log_8 x = \frac{\log_2 x}{\log_2 8} = \frac{\log_2 x}{3}$ since $\log_2 8 = 3$.\n\n4. **Substitute and simplify:**\n$$\log_2 x + \frac{\log_2 x}{2} + \frac{\log_2 x}{3} = \frac{22}{3}$$\nLet $y = \log_2 x$. Then,\n$$y + \frac{y}{2} + \frac{y}{3} = \frac{22}{3}$$\n\n5. **Combine like terms:**\nFind common denominator 6:\n$$\frac{6y}{6} + \frac{3y}{6} + \frac{2y}{6} = \frac{22}{3}$$\n$$\frac{6y + 3y + 2y}{6} = \frac{22}{3}$$\n$$\frac{11y}{6} = \frac{22}{3}$$\n\n6. **Solve for $y$:**\nMultiply both sides by 6:\n$$11y = \frac{22}{3} \times 6 = 22 \times 2 = 44$$\n$$y = \frac{44}{11} = 4$$\n\n7. **Back-substitute to find $x$:**\nRecall $y = \log_2 x = 4$, so\n$$x = 2^4 = 16$$\n\n**Final answer:** $x = 16$