1. **State the problem:** Solve the equation $$4^x - 13 \times 2^x + 40 = 0$$. 2. **Rewrite the equation using powers of 2:** Note that $$4^x = (2^2)^x = 2^{2x} = (2^x)^2$$. 3. **Substitute:** Let $$y = 2^x$$, so the equation becomes $$y^2 - 13y + 40 = 0$$. 4. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=1$$, $$b=-13$$, and $$c=40$$. 5. **Calculate the discriminant:** $$\Delta = (-13)^2 - 4 \times 1 \times 40 = 169 - 160 = 9$$. 6. **Find the roots:** $$y = \frac{13 \pm \sqrt{9}}{2} = \frac{13 \pm 3}{2}$$ 7. **Evaluate each root:** - $$y_1 = \frac{13 + 3}{2} = \frac{16}{2} = 8$$ - $$y_2 = \frac{13 - 3}{2} = \frac{10}{2} = 5$$ 8. **Back-substitute:** Recall $$y = 2^x$$, so solve for $$x$$: - For $$2^x = 8$$, since $$8 = 2^3$$, $$x = 3$$. - For $$2^x = 5$$, take logarithm base 2: $$x = \log_2 5 = \frac{\ln 5}{\ln 2}$$. **Final answer:** $$x = 3$$ or $$x = \frac{\ln 5}{\ln 2}$$.