(11) Given that $a, b, c, d$ are all integers, and $\log _{a} b=\frac{3}{2} \cdot \log _{c} d=\frac{5}{4}$. If $a-c=9$, then $b-d=$ $\qquad$ .
From the conditions, we know that $b=a^{\frac{3}{2}}, d=c^{\frac{1}{4}}$. Since $a, b, c, d$ are all positive integers, assume $a=x^{2}, c=y^{4}$, where $x, y$ are positive integers. Using $a-c=9$, we get $\left(x-y^{2}\right)\left(x+y^{2}\right)=9$, which can only be $\left(x-y^{2}, x+y^{2}\right)=(1,9)$. Solving this, we get $x=5 \cdot y=2$. Therefore, $b-d=x^{3}-y^{5}=125-32=93$.