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:** - $\sqrt{2} = 2^{\frac{1}{2}}$. - For any base $a$ and exponents $m,n$, $a^{m+n} = a^m \cdot a^n$. - Also, $a^{m-n} = \frac{a^m}{a^n}$. 3. **Rewrite the terms using base 2:** $$ (\sqrt{2})^{4+x} = (2^{\frac{1}{2}})^{4+x} = 2^{\frac{4+x}{2}} $$ $$ (\sqrt{2})^{2-x} = (2^{\frac{1}{2}})^{2-x} = 2^{\frac{2-x}{2}} $$ 4. **Rewrite the right side:** $$ 3\sqrt{2} = 3 \cdot 2^{\frac{1}{2}} $$ 5. **Substitute back into the equation:** $$ 2^{\frac{4+x}{2}} + 2^{\frac{2-x}{2}} = 3 \cdot 2^{\frac{1}{2}} $$ 6. **Let $y = 2^{\frac{x}{2}}$. Then:** $$ 2^{\frac{4}{2}} \cdot 2^{\frac{x}{2}} + 2^{\frac{2}{2}} \cdot 2^{-\frac{x}{2}} = 3 \cdot 2^{\frac{1}{2}} $$ $$ 2^2 \cdot y + 2^1 \cdot \frac{1}{y} = 3 \cdot 2^{\frac{1}{2}} $$ $$ 4y + \frac{2}{y} = 3 \sqrt{2} $$ 7. **Multiply both sides by $y$ to clear the denominator:** $$ 4y^2 + 2 = 3 \sqrt{2} y $$ 8. **Rearrange into a quadratic form:** $$ 4y^2 - 3 \sqrt{2} y + 2 = 0 $$ 9. **Use the quadratic formula:** $$ y = \frac{3 \sqrt{2} \pm \sqrt{(3 \sqrt{2})^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} $$ 10. **Calculate the discriminant:** $$ (3 \sqrt{2})^2 = 9 \cdot 2 = 18 $$ $$ 4 \cdot 4 \cdot 2 = 32 $$ $$ \text{Discriminant} = 18 - 32 = -14 $$ 11. **Since the discriminant is negative, there are no real solutions for $y$.** 12. **Check for possible mistakes or consider complex solutions:** - The problem likely expects real solutions, so let's re-express the original equation differently. 13. **Alternative substitution:** Rewrite the original equation as: $$ (\sqrt{2})^{4+x} + (\sqrt{2})^{2-x} = 3 \sqrt{2} $$ Let $a = \sqrt{2}$. 14. **Express terms:** $$ a^{4+x} = a^4 \cdot a^x $$ $$ a^{2-x} = a^2 \cdot a^{-x} $$ 15. **Let $t = a^x$, then:** $$ a^4 t + a^2 \frac{1}{t} = 3 a $$ 16. **Multiply both sides by $t$:** $$ a^4 t^2 + a^2 = 3 a t $$ 17. **Rearranged quadratic in $t$:** $$ a^4 t^2 - 3 a t + a^2 = 0 $$ 18. **Substitute $a = \sqrt{2} = 2^{1/2}$:** $$ a^4 = (2^{1/2})^4 = 2^{2} = 4 $$ $$ a^2 = 2 $$ $$ a = \sqrt{2} $$ 19. **Quadratic becomes:** $$ 4 t^2 - 3 \sqrt{2} t + 2 = 0 $$ 20. **Calculate discriminant:** $$ D = ( -3 \sqrt{2} )^2 - 4 \cdot 4 \cdot 2 = 18 - 32 = -14 $$ 21. **Again, discriminant is negative, no real solutions for $t$.** 22. **Conclusion:** The equation has no real solutions for $x$ because the quadratic in $t = (\sqrt{2})^x$ has no real roots. **Final answer:** $$ \boxed{\text{No real solution for } x} $$