1. **State the problem:** We are given the inequalities $0 < x < y^2 < x^2$ and $y > 1$. We need to sort the expressions $A = \log(y) y \sqrt{x}$, $B = \log(x) \frac{x^2}{y}$, $C = \log(x) y$, and $D = \log(y) x$ in ascending order. 2. **Analyze the inequalities:** Since $y > 1$, $\log(y) > 0$ because the logarithm of a number greater than 1 is positive. 3. From $0 < x < y^2 < x^2$, we deduce: - $x < y^2$ and - $y^2 < x^2$ which implies $y < x$ (since both $x$ and $y$ are positive). 4. Since $y > 1$ and $y < x$, it follows that $x > y > 1$. 5. Because $x > 1$, $\log(x) > 0$ as well. 6. Now evaluate each expression qualitatively: - $A = \log(y) y \sqrt{x}$: positive factors multiplied. - $B = \log(x) \frac{x^2}{y}$: positive factors, $x^2/y$ is large since $x > y$. - $C = \log(x) y$: positive factors. - $D = \log(y) x$: positive factors. 7. Compare $A$ and $D$: - $A = \log(y) y \sqrt{x}$ - $D = \log(y) x$ Since $x > \sqrt{x}$ and $y > 1$, $D > A$ because $x > y \sqrt{x}$. 8. Compare $B$ and $C$: - $B = \log(x) \frac{x^2}{y}$ - $C = \log(x) y$ Since $x^2/y > y$ (because $x > y$), $B > C$. 9. Compare $C$ and $A$: - $C = \log(x) y$ - $A = \log(y) y \sqrt{x}$ Since $\log(x) > \log(y)$ and $\sqrt{x} < x$, but $A$ has $\sqrt{x}$ and $C$ has $y$, and $y < x$, this is less straightforward. But since $\log(x) > \log(y)$ and $y$ is common, $C > A$. 10. Compare $C$ and $D$: - $C = \log(x) y$ - $D = \log(y) x$ Since $\log(x) > \log(y)$ and $x > y$, the comparison depends on the relative sizes, but generally $C > D$ because $\log(x)$ grows faster. 11. Final ascending order: $$ A < D < C < B $$ **Answer:** The expressions in ascending order are $A$, $D$, $C$, $B$.