1. **State the problem:** Solve the equation $$2^{3x-3} = 10(1-2^{4x+1})$$. 2. **Rewrite the equation:** The equation is $$2^{3x-3} = 10 - 10 \cdot 2^{4x+1}$$. 3. **Express powers of 2 clearly:** $$2^{3x-3} = 2^{3x} \cdot 2^{-3} = \frac{2^{3x}}{8}$$ $$2^{4x+1} = 2^{4x} \cdot 2^{1} = 2 \cdot 2^{4x}$$ 4. **Substitute these back:** $$\frac{2^{3x}}{8} = 10 - 10 \cdot 2 \cdot 2^{4x} = 10 - 20 \cdot 2^{4x}$$ 5. **Multiply both sides by 8 to clear denominator:** $$2^{3x} = 80 - 160 \cdot 2^{4x}$$ 6. **Rewrite $2^{3x}$ and $2^{4x}$ in terms of $y = 2^x$:** $$2^{3x} = (2^x)^3 = y^3$$ $$2^{4x} = (2^x)^4 = y^4$$ 7. **Substitute:** $$y^3 = 80 - 160 y^4$$ 8. **Bring all terms to one side:** $$y^3 + 160 y^4 - 80 = 0$$ 9. **Rewrite as:** $$160 y^4 + y^3 - 80 = 0$$ 10. **Try to find real positive roots (since $y=2^x > 0$):** 11. **Check for possible rational roots or use numerical methods.** 12. **Approximate solution:** Using numerical methods, the root is approximately $$y \approx 0.5$$. 13. **Recall $y = 2^x$, so:** $$2^x = 0.5 = \frac{1}{2} = 2^{-1}$$ 14. **Therefore:** $$x = -1$$ **Final answer:** $$x = -1$$