1. **State the problem:** Solve the equation $2^x = 7 + 2^{2-x}$ for $x$. 2. **Rewrite the equation:** The equation is $2^x = 7 + 2^{2-x}$. 3. **Express $2^{2-x}$ in terms of $2^x$: ** Recall that $2^{2-x} = 2^2 \cdot 2^{-x} = 4 \cdot \frac{1}{2^x} = \frac{4}{2^x}$. 4. **Substitute into the equation:** $$2^x = 7 + \frac{4}{2^x}$$ 5. **Multiply both sides by $2^x$ to clear the denominator:** $$\cancel{2^x} \cdot 2^x = 7 \cdot \cancel{2^x} + 4$$ $$ (2^x)^2 = 7 \cdot 2^x + 4$$ 6. **Let $y = 2^x$, then the equation becomes:** $$y^2 = 7y + 4$$ 7. **Rewrite as a quadratic equation:** $$y^2 - 7y - 4 = 0$$ 8. **Use the quadratic formula to solve for $y$:** $$y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} = \frac{7 \pm \sqrt{49 + 16}}{2} = \frac{7 \pm \sqrt{65}}{2}$$ 9. **Calculate the approximate values:** $$\sqrt{65} \approx 8.062$$ So, $$y_1 = \frac{7 + 8.062}{2} = \frac{15.062}{2} = 7.531$$ $$y_2 = \frac{7 - 8.062}{2} = \frac{-1.062}{2} = -0.531$$ 10. **Since $y = 2^x$ and $2^x > 0$ for all real $x$, discard $y_2 = -0.531$ (negative).** 11. **Solve for $x$ using $y_1 = 7.531$:** $$2^x = 7.531$$ Take logarithm base 2 on both sides: $$x = \log_2(7.531) = \frac{\ln(7.531)}{\ln(2)}$$ 12. **Calculate the value:** $$\ln(7.531) \approx 2.020$$ $$\ln(2) \approx 0.693$$ $$x \approx \frac{2.020}{0.693} \approx 2.915$$ **Final answer:** $$x \approx 2.915$$