1. **State the problem:** Given the equation $\log \sqrt{x} = -1$, find the value of $x$. 2. **Recall the logarithm and root properties:** - $\sqrt{x} = x^{\frac{1}{2}}$ - $\log a^b = b \log a$ 3. **Rewrite the equation using these properties:** $$\log \sqrt{x} = \log x^{\frac{1}{2}} = \frac{1}{2} \log x$$ So the equation becomes: $$\frac{1}{2} \log x = -1$$ 4. **Solve for $\log x$:** Multiply both sides by 2: $$\log x = -2$$ 5. **Convert from logarithmic to exponential form:** Recall that $\log x = y$ means $10^y = x$. So, $$x = 10^{-2}$$ 6. **Evaluate the options:** - a. $0.001 = 10^{-3}$ (incorrect) - b. $0.1 = 10^{-1}$ (incorrect) - c. $10^{-2}$ (correct) - d. $-0.01$ (negative, invalid since $x$ must be positive for $\log x$) **Final answer:** $x = 10^{-2}$ (option c)