1. **Problem Statement:** Solve for $x$ in the equation $x + \log_2(2^x - 6) = 4$. 2. **Formula and Rules:** Recall that $\log_b(a)$ is the logarithm base $b$ of $a$, and the domain requires $2^x - 6 > 0$ so that the logarithm is defined. 3. **Step 1: Domain restriction** $$2^x - 6 > 0 \implies 2^x > 6$$ 4. **Step 2: Rewrite the equation** $$x + \log_2(2^x - 6) = 4$$ 5. **Step 3: Let $y = 2^x$** Then $x = \log_2 y$, so substitute: $$\log_2 y + \log_2(y - 6) = 4$$ 6. **Step 4: Use logarithm property** $$\log_2[y(y - 6)] = 4$$ 7. **Step 5: Convert logarithmic to exponential form** $$y(y - 6) = 2^4 = 16$$ 8. **Step 6: Form quadratic equation** $$y^2 - 6y - 16 = 0$$ 9. **Step 7: Solve quadratic equation** $$y = \frac{6 \pm \sqrt{36 + 64}}{2} = \frac{6 \pm 10}{2}$$ 10. **Step 8: Find roots** - $y = \frac{6 + 10}{2} = 8$ - $y = \frac{6 - 10}{2} = -2$ (discard since $y=2^x > 0$) 11. **Step 9: Back-substitute for $x$** $$2^x = 8 \implies x = \log_2 8 = 3$$ 12. **Step 10: Verify domain** Since $2^3 - 6 = 8 - 6 = 2 > 0$, the solution is valid. **Final answer:** $$\boxed{3}$$