Yet another log

$\begin{cases} \log_a b = \frac{3}{2} & \Rightarrow b = a^{\frac{3}{2}} & ...(1) \\ \log_c d = \frac{5}{2} & \Rightarrow d = c^{\frac{5}{2}} & ...(2) \\ a - c = 9 & \Rightarrow a = c+9 & ...(3) \end{cases}$

$\begin{aligned} b - d & = \color{#3D99F6}{a^\frac{3}{2}} - c^\frac{5}{4} \quad \quad \quad \quad \small \color{#3D99F6}{\text {Substituting }(3): a = c+9} \\ & = \color{#3D99F6}{(c+9)^\frac{3}{2}} - c^\frac{5}{4} \end{aligned}$

For integer solution, $(c+9)^\frac{1}{2}$ and $c^\frac{1}{4}$ must be integers. Therefore, $c$ is a perfect square; let $c = n^2$ . And $c+9 = n^2+3^2$ is also a perfect square. We note that if $n$ is a Pythagorean triple, it will fit the bill. In this case, $n=4$ and $c=16 = 2^4$ .

Therefore we have:

$\begin{aligned} b - d & = (c+9)^\frac{3}{2} - c^\frac{5}{4} \\ & = 5^3 - 2^5 = 125 - 32 = \boxed{93} \end{aligned}$

$a-c=9$ : $b^{2/3}-d^{4/5}=9: (b^{1/3}+d^{2/5}) \times (b^{1/3}-d^{2/5})=9= 9 \times 1$ ( first bracket is bigger).
Assume A= $b^{1/3}$ and B= $d^{2/5}$ . solving 2 variable 2 equation, $A=5$ and $B=4 => b=125,d=32,b-d=93$
One more way is to think that 9 is only the difference of 2 perfect squares i.e. 25-16. Put the values and answer is same. Notice that i have not written 9=9-0 since none of them could be 0 as log won't be defined. Clarification: $9=1 \times 9= 3 \times 3= 9 \times 1$ out of which only last is valid because $b^{1/3}+d^{2/5}> b^{1/3}-d^{2/5}$ . If they both are equal, then $d=0$ which is not valid.