1. **Stating the problem:** Solve the equation $$x^x = 9\sqrt{\frac{1}{3}}$$ for $x$. 2. **Rewrite the right side:** Note that $$9\sqrt{\frac{1}{3}} = 9 \times \left(\frac{1}{3}\right)^{\frac{1}{2}} = 9 \times 3^{-\frac{1}{2}} = 9 \times \frac{1}{\sqrt{3}}.$$ Simplify this: $$9 \times \frac{1}{\sqrt{3}} = \frac{9}{\sqrt{3}} = 3 \times \frac{3}{\sqrt{3}} = 3 \times \sqrt{3} = 3^{1} \times 3^{\frac{1}{2}} = 3^{\frac{3}{2}}.$$ So, $$x^x = 3^{\frac{3}{2}}.$$ 3. **Express $x$ as a power of 3:** Let $$x = 3^a$$ for some real number $a$. 4. **Rewrite the equation using $a$:** $$x^x = (3^a)^{3^a} = 3^{a \cdot 3^a} = 3^{\frac{3}{2}}.$$ 5. **Equate exponents:** Since the bases are the same (3), the exponents must be equal: $$a \cdot 3^a = \frac{3}{2}.$$ 6. **Solve for $a$:** This is a transcendental equation and generally requires numerical methods. However, try to find an integer or simple rational solution: - For $a=\frac{1}{2}$, $$a \cdot 3^a = \frac{1}{2} \times 3^{\frac{1}{2}} = \frac{1}{2} \times \sqrt{3} \approx 0.866,$$ which is less than $1.5$. - For $a=1$, $$1 \times 3^1 = 3,$$ which is greater than $1.5$. So $a$ is between $0.5$ and $1$. 7. **Approximate $a$ numerically:** Using trial, - For $a=0.7$, $$0.7 \times 3^{0.7} \approx 0.7 \times 2.039 = 1.427,$$ - For $a=0.75$, $$0.75 \times 3^{0.75} \approx 0.75 \times 2.279 = 1.709,$$ So $a \approx 0.7$ to $0.75$. More precise numerical methods (like Newton-Raphson) can find $a \approx 0.72$. 8. **Final answer:** $$x = 3^a \approx 3^{0.72} \approx 2.17.$$