1. **State the problem:** Solve the exponential equation $$2^{2x} - 2^x - 12 = 0$$ for $x$. 2. **Rewrite the equation:** Notice that $2^{2x} = (2^x)^2$. Let $y = 2^x$. Then the equation becomes: $$y^2 - y - 12 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-1$, and $c=-12$. 4. **Calculate the discriminant:** $$\sqrt{(-1)^2 - 4 \times 1 \times (-12)} = \sqrt{1 + 48} = \sqrt{49} = 7$$ 5. **Find the roots:** $$y = \frac{1 \pm 7}{2}$$ 6. **Evaluate each root:** - For $y = \frac{1 + 7}{2} = \frac{8}{2} = 4$ - For $y = \frac{1 - 7}{2} = \frac{-6}{2} = -3$ 7. **Check for valid solutions:** Since $y = 2^x$ and $2^x > 0$ for all real $x$, discard $y = -3$. 8. **Solve for $x$:** $$2^x = 4$$ Since $4 = 2^2$, we have: $$2^x = 2^2$$ Therefore: $$x = 2$$ **Final answer:** $$x = 2$$