1. The problem states: Given $a>b>c>1$ and $b^2=ac$, sort $A=\log_b c$, $B=\log_c a$, and $C=\log_a b$ in ascending order. 2. Recall the change of base formula for logarithms: $\log_x y = \frac{\log y}{\log x}$ where the base of the logarithm on the right side is any positive number (commonly 10 or $e$). 3. Express $A$, $B$, and $C$ using natural logarithms $\ln$: $$A = \log_b c = \frac{\ln c}{\ln b}, \quad B = \log_c a = \frac{\ln a}{\ln c}, \quad C = \log_a b = \frac{\ln b}{\ln a}$$ 4. Given $b^2 = ac$, take natural logarithms: $$2 \ln b = \ln a + \ln c$$ 5. Let $x = \ln a$, $y = \ln b$, $z = \ln c$. Then $2y = x + z$. 6. Since $a > b > c > 1$, we have $x > y > z > 0$. 7. Rewrite $A$, $B$, and $C$ in terms of $x,y,z$: $$A = \frac{z}{y}, \quad B = \frac{x}{z}, \quad C = \frac{y}{x}$$ 8. From $2y = x + z$, we get $y = \frac{x+z}{2}$. 9. Compare $A$ and $C$: $$A - C = \frac{z}{y} - \frac{y}{x} = \frac{z x - y^2}{y x}$$ 10. Substitute $y = \frac{x+z}{2}$: $$z x - y^2 = z x - \left(\frac{x+z}{2}\right)^2 = z x - \frac{x^2 + 2 x z + z^2}{4} = \frac{4 z x - x^2 - 2 x z - z^2}{4} = \frac{2 z x - x^2 - z^2}{4}$$ 11. Rewrite numerator: $$2 z x - x^2 - z^2 = -(x - z)^2 < 0$$ 12. Since denominator $y x > 0$, $A - C < 0$, so $A < C$. 13. Compare $B$ and $C$: $$B - C = \frac{x}{z} - \frac{y}{x} = \frac{x^2 - y z}{z x}$$ 14. Substitute $y = \frac{x+z}{2}$: $$x^2 - y z = x^2 - \frac{x+z}{2} z = x^2 - \frac{x z + z^2}{2} = \frac{2 x^2 - x z - z^2}{2}$$ 15. Since $x > z > 0$, $2 x^2 - x z - z^2 > 0$, so $B - C > 0$, thus $B > C$. 16. Compare $A$ and $B$: $$B - A = \frac{x}{z} - \frac{z}{y} = \frac{x y - z^2}{z y}$$ 17. Substitute $y = \frac{x+z}{2}$: $$x y - z^2 = x \frac{x+z}{2} - z^2 = \frac{x^2 + x z - 2 z^2}{2}$$ 18. Since $x > z$, $x^2 + x z - 2 z^2 > 0$, so $B - A > 0$, thus $B > A$. 19. Summary of inequalities: $$A < C < B$$ 20. Therefore, the ascending order is $A$, then $C$, then $B$.