1. The problem is to solve the equation $\log x^{\log x} = 4$. 2. Recall the logarithm power rule: $\log a^b = b \log a$. 3. Apply this rule to the left side: $$\log x^{\log x} = (\log x)(\log x) = (\log x)^2$$ 4. So the equation becomes: $$(\log x)^2 = 4$$ 5. Take the square root of both sides: $$\log x = \pm 2$$ 6. Solve for $x$ by rewriting the logarithm in exponential form: - If $\log x = 2$, then $x = 10^2 = 100$. - If $\log x = -2$, then $x = 10^{-2} = 0.01$. 7. Both values are valid since $x > 0$ for logarithms. **Final answer:** $$x = 100 \text{ or } x = 0.01$$