1. **State the problem:** Solve the equation $$ (\sqrt{2})^{4+x} + (\sqrt{2})^{2 - x} = 3\sqrt{2} $$ for $x$. 2. **Recall the properties of exponents:** - $\sqrt{2} = 2^{1/2}$. - When raising a power to another power, multiply exponents: $\left(a^m\right)^n = a^{mn}$. 3. **Rewrite the bases:** $$ (\sqrt{2})^{4+x} = \left(2^{1/2}\right)^{4+x} = 2^{\frac{4+x}{2}} $$ $$ (\sqrt{2})^{2-x} = \left(2^{1/2}\right)^{2-x} = 2^{\frac{2-x}{2}} $$ 4. **Substitute into the equation:** $$ 2^{\frac{4+x}{2}} + 2^{\frac{2-x}{2}} = 3 \times 2^{1/2} $$ 5. **Let $y = 2^{x/2}$ to simplify:** - Then $2^{\frac{4+x}{2}} = 2^2 \times 2^{x/2} = 4y$ - And $2^{\frac{2-x}{2}} = 2^1 \times 2^{-x/2} = 2 \times \frac{1}{y} = \frac{2}{y}$ 6. **Rewrite the equation in terms of $y$:** $$ 4y + \frac{2}{y} = 3 \times 2^{1/2} $$ 7. **Multiply both sides by $y$ to clear the denominator:** $$ 4y^2 + 2 = 3 \sqrt{2} y $$ 8. **Rearrange to form a quadratic equation:** $$ 4y^2 - 3 \sqrt{2} y + 2 = 0 $$ 9. **Use the quadratic formula:** $$ y = \frac{3 \sqrt{2} \pm \sqrt{(3 \sqrt{2})^2 - 4 \times 4 \times 2}}{2 \times 4} $$ Calculate the discriminant: $$ (3 \sqrt{2})^2 = 9 \times 2 = 18 $$ $$ 4 \times 4 \times 2 = 32 $$ $$ \text{Discriminant} = 18 - 32 = -14 $$ 10. **Since the discriminant is negative, there are no real solutions for $y$.** 11. **Check for possible mistakes or consider complex solutions:** - The problem likely expects real solutions, so let's re-express the original equation differently. 12. **Alternative approach: Express $\sqrt{2}$ as $2^{1/2}$ and rewrite the original equation:** $$ (2^{1/2})^{4+x} + (2^{1/2})^{2-x} = 3 \times 2^{1/2} $$ $$ 2^{\frac{4+x}{2}} + 2^{\frac{2-x}{2}} = 3 \times 2^{1/2} $$ 13. **Let $a = 2^{x/2}$, then:** $$ 2^{2} \times a + 2^{1} \times \frac{1}{a} = 3 \times 2^{1/2} $$ $$ 4a + \frac{2}{a} = 3 \sqrt{2} $$ 14. **Multiply both sides by $a$:** $$ 4a^2 + 2 = 3 \sqrt{2} a $$ 15. **Rearranged quadratic:** $$ 4a^2 - 3 \sqrt{2} a + 2 = 0 $$ 16. **Discriminant again:** $$ (3 \sqrt{2})^2 - 4 \times 4 \times 2 = 18 - 32 = -14 $$ 17. **No real solutions for $a$, so no real $x$.** 18. **Conclusion:** The equation has no real solutions for $x$ because the quadratic in $a$ has a negative discriminant. **Final answer:** No real solution for $x$.