1. **Problem a:** Simplify $\left(\frac{a}{b}\right)^{\log 0.5} \cdot \left(\frac{a}{b}\right)^{\log 0.2}$. 2. Use the property of exponents: $x^m \cdot x^n = x^{m+n}$. 3. Combine the exponents: $$\left(\frac{a}{b}\right)^{\log 0.5 + \log 0.2}$$ 4. Use the logarithm property: $\log m + \log n = \log(m \cdot n)$. 5. So the exponent becomes: $$\log(0.5 \times 0.2) = \log 0.1$$ 6. Therefore, the expression is: $$\left(\frac{a}{b}\right)^{\log 0.1}$$ 7. **Problem b:** Simplify $\frac{\log_a x}{\log_{ab} x} - \frac{\log_a x}{\log_b x}$. 8. Recall the change of base formula: $\log_c d = \frac{\log d}{\log c}$ for any base. 9. Express all logs in a common base, say base 10: $$\log_a x = \frac{\log x}{\log a}, \quad \log_b x = \frac{\log x}{\log b}, \quad \log_{ab} x = \frac{\log x}{\log(ab)} = \frac{\log x}{\log a + \log b}$$ 10. Substitute into the expression: $$\frac{\frac{\log x}{\log a}}{\frac{\log x}{\log a + \log b}} - \frac{\frac{\log x}{\log a}}{\frac{\log x}{\log b}}$$ 11. Simplify each fraction: $$\frac{\log x}{\log a} \cdot \frac{\log a + \log b}{\log x} - \frac{\log x}{\log a} \cdot \frac{\log b}{\log x}$$ 12. Cancel $\log x$ where possible: $$\cancel{\log x} \cdot \frac{\log a + \log b}{\log a} \cdot \frac{1}{\cancel{\log x}} - \cancel{\log x} \cdot \frac{\log b}{\log a} \cdot \frac{1}{\cancel{\log x}}$$ 13. This simplifies to: $$\frac{\log a + \log b}{\log a} - \frac{\log b}{\log a}$$ 14. Split the first fraction: $$\frac{\log a}{\log a} + \frac{\log b}{\log a} - \frac{\log b}{\log a}$$ 15. Cancel terms: $$1 + \frac{\log b}{\log a} - \frac{\log b}{\log a} = 1$$ **Final answers:** - a) $\boxed{\left(\frac{a}{b}\right)^{\log 0.1}}$ - b) $\boxed{1}$