<s>[INST] If $a-c = 9$ then find the value of $b-d$. If $a,b,c,d$ are positive integers such that $\log_a b=\frac{3}{2}$ and $\log_c d=\frac{5}{4}$, if $a-c = 9$ then find the value of $b-d$.
We get
$b=a^{3/2}$ and $d=c^{5/4}$
Hence
$b-d=a^{3/2}-c^{5/4}$
Now we need to use the fact that $a,b,c,d$ are integers, but is hit and trial the only way to go here?
 [/INST] For $b$ and $d$ to be integers, we need some integers $p$ and $q$ such that $p^2=a$ and $q^4=c$.
So, $p^2-q^4=9$, and we want to find $p^3-q^5$. 
$(p+q^2)(p-q^2)=9$. As $p$ and $q$ are integers, so must $(p+q^2)$ and $(p-q^2)$. Considering their product is 9, we have either: $p+q^2=9$ and $p-q^2=1$, or both are equal to $3$. The latter event locks $q=0$, meaning $c=0$, which means that $log_c d$ is undefined, so it must be the former.
$(p+q^2)-(p-q^2)=9-1$ reduces to $2q^2=8$, meaning $q=2$. From there, it is easy to see that $p=5$.
$b=p^3=125$; $d=q^5=32$. $b-d=125-32=93$.
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