1. **State the problem:** Solve the equation $x^x = 2^{10\sqrt{200}}$ for $x$. 2. **Rewrite the right side:** Simplify the exponent on the right side. $$10\sqrt{200} = 10 \times \sqrt{100 \times 2} = 10 \times 10 \sqrt{2} = 100\sqrt{2}$$ So the equation becomes: $$x^x = 2^{100\sqrt{2}}$$ 3. **Take the natural logarithm of both sides:** $$\ln(x^x) = \ln\left(2^{100\sqrt{2}}\right)$$ Using the logarithm power rule $\ln(a^b) = b \ln(a)$: $$x \ln(x) = 100\sqrt{2} \ln(2)$$ 4. **Analyze the equation:** We want to find $x$ such that: $$x \ln(x) = 100\sqrt{2} \ln(2)$$ 5. **Try to express $x$ as a power of 2:** Let $x = 2^k$. Then: $$x \ln(x) = 2^k \ln(2^k) = 2^k \times k \ln(2) = k 2^k \ln(2)$$ Set equal to the right side: $$k 2^k \ln(2) = 100\sqrt{2} \ln(2)$$ Divide both sides by $\ln(2)$: $$k 2^k = 100\sqrt{2}$$ 6. **Solve for $k$:** We need $k$ such that: $$k 2^k = 100\sqrt{2}$$ This is a transcendental equation and can be solved numerically. 7. **Approximate $k$:** Try $k=6$: $$6 \times 2^6 = 6 \times 64 = 384$$ Try $k=7$: $$7 \times 2^7 = 7 \times 128 = 896$$ Try $k=8$: $$8 \times 2^8 = 8 \times 256 = 2048$$ Try $k=5$: $$5 \times 2^5 = 5 \times 32 = 160$$ Try $k=4$: $$4 \times 2^4 = 4 \times 16 = 64$$ Since $100\sqrt{2} \approx 141.42$, $k$ is between 4 and 5. Try $k=4.5$: $$4.5 \times 2^{4.5} = 4.5 \times 2^{4} \times 2^{0.5} = 4.5 \times 16 \times \sqrt{2} = 4.5 \times 16 \times 1.4142 = 4.5 \times 22.627 = 101.82$$ Close to 141.42, try $k=4.8$: $$4.8 \times 2^{4.8} = 4.8 \times 2^{4} \times 2^{0.8} = 4.8 \times 16 \times 2^{0.8}$$ Calculate $2^{0.8} = e^{0.8 \ln(2)} = e^{0.8 \times 0.6931} = e^{0.5545} \approx 1.7408$ So: $$4.8 \times 16 \times 1.7408 = 4.8 \times 27.8528 = 133.7$$ Try $k=5$ again (160) and $k=4.9$: $$4.9 \times 16 \times 2^{0.9}$$ Calculate $2^{0.9} = e^{0.9 \times 0.6931} = e^{0.6238} \approx 1.865$$ So: $$4.9 \times 16 \times 1.865 = 4.9 \times 29.84 = 146.2$$ Since $141.42$ is between $133.7$ and $146.2$, $k \approx 4.85$. 8. **Final answer:** $$x = 2^{k} \approx 2^{4.85}$$ Calculate: $$2^{4.85} = 2^{4} \times 2^{0.85} = 16 \times e^{0.85 \times 0.6931} = 16 \times e^{0.589} = 16 \times 1.802 = 28.83$$ **Therefore, the solution is approximately:** $$\boxed{x \approx 28.83}$$