1. **State the problem:** Solve for $x$ in the equation $$\log_{2^x} 16 + 3 \log_{3^x} 8 = 1.$$\n\n2. **Recall the logarithm change of base formula:** For any positive $a,b,c$ with $a \neq 1$, $$\log_a b = \frac{\log b}{\log a}.$$\n\n3. **Rewrite each logarithm using the change of base formula:**\n$$\log_{2^x} 16 = \frac{\log 16}{\log 2^x} = \frac{\log 16}{x \log 2}$$\n$$\log_{3^x} 8 = \frac{\log 8}{\log 3^x} = \frac{\log 8}{x \log 3}.$$\n\n4. **Substitute these back into the equation:**\n$$\frac{\log 16}{x \log 2} + 3 \cdot \frac{\log 8}{x \log 3} = 1.$$\n\n5. **Simplify the logarithms of powers:**\nSince $16 = 2^4$, $\log 16 = 4 \log 2$.\nSince $8 = 2^3$, $\log 8 = 3 \log 2$.\n\n6. **Substitute these values:**\n$$\frac{4 \log 2}{x \log 2} + 3 \cdot \frac{3 \log 2}{x \log 3} = 1.$$\n\n7. **Simplify the fractions:**\n$$\frac{4 \cancel{\log 2}}{x \cancel{\log 2}} + 3 \cdot \frac{3 \log 2}{x \log 3} = \frac{4}{x} + \frac{9 \log 2}{x \log 3} = 1.$$\n\n8. **Combine terms over common denominator $x$:**\n$$\frac{4 + 9 \frac{\log 2}{\log 3}}{x} = 1.$$\n\n9. **Multiply both sides by $x$:**\n$$4 + 9 \frac{\log 2}{\log 3} = x.$$\n\n10. **Calculate the numerical value:**\nUsing natural logs or common logs, $\frac{\log 2}{\log 3} \approx 0.6309$, so\n$$x \approx 4 + 9 \times 0.6309 = 4 + 5.6781 = 9.6781.$$\n\n**Final answer:** $$x = 4 + 9 \frac{\log 2}{\log 3} \approx 9.678.$$