1. **State the problem:** Solve the equation $$\log_4 x^{\log_4 729} - \log_3 27^{\log_4 x} = 15.$$\n\n2. **Recall logarithm rules:**\n- Power rule: $$\log_a b^c = c \log_a b.$$\n- Change of base formula: $$\log_a b = \frac{\log_c b}{\log_c a}$$ for any positive base $c \neq 1$.\n\n3. **Rewrite each term using the power rule:**\n$$\log_4 x^{\log_4 729} = (\log_4 729)(\log_4 x)$$\n$$\log_3 27^{\log_4 x} = (\log_4 x)(\log_3 27)$$\n\n4. **Substitute these back into the equation:**\n$$(\log_4 729)(\log_4 x) - (\log_4 x)(\log_3 27) = 15.$$\n\n5. **Factor out $\log_4 x$:**\n$$\log_4 x \left( \log_4 729 - \log_3 27 \right) = 15.$$\n\n6. **Evaluate the logarithms:**\n- $729 = 3^6$, so $$\log_4 729 = \log_4 3^6 = 6 \log_4 3.$$\n- $27 = 3^3$, so $$\log_3 27 = \log_3 3^3 = 3.$$\n\n7. **Substitute these values:**\n$$\log_4 x (6 \log_4 3 - 3) = 15.$$\n\n8. **Use change of base to express $\log_4 3$ in terms of $\log_3 4$ or vice versa, or approximate:**\nRecall $$\log_4 3 = \frac{\log 3}{\log 4}$$ (common log or natural log).\n\n9. **Rewrite the expression inside parentheses:**\n$$6 \log_4 3 - 3 = 3(2 \log_4 3 - 1).$$\n\n10. **Set $y = \log_4 x$ for simplicity:**\n$$y \cdot 3(2 \log_4 3 - 1) = 15 \implies y = \frac{15}{3(2 \log_4 3 - 1)} = \frac{5}{2 \log_4 3 - 1}.$$\n\n11. **Calculate $\log_4 3$ approximately:**\nUsing natural logs, $$\log_4 3 = \frac{\ln 3}{\ln 4} \approx \frac{1.0986}{1.3863} \approx 0.7925.$$\n\n12. **Calculate denominator:**\n$$2 \times 0.7925 - 1 = 1.585 - 1 = 0.585.$$\n\n13. **Calculate $y$:**\n$$y = \frac{5}{0.585} \approx 8.547.$$\n\n14. **Recall $y = \log_4 x$, so:**\n$$\log_4 x = 8.547 \implies x = 4^{8.547}.$$\n\n15. **Calculate $x$ approximately:**\n$$x \approx 4^{8.547} = e^{8.547 \ln 4} \approx e^{8.547 \times 1.3863} = e^{11.85} \approx 1.4 \times 10^5.$$\n\n**Final answer:** $$x \approx 1.4 \times 10^5.$$