Find $b-d$ when $\log_ab={3\over2}$ and $\log_cd={5\over4}$ $a,b,c$ are three natural numbers such that $\log_ab={3\over2}$ and $\log_cd={5\over4}$.
Given: $a-c=9$
Find $b-d$

Hint: $$\log_ab={3\over2} \Rightarrow b=a^{\frac32}\Rightarrow a \text{ is a perfect square, } a= \alpha^2 \to b=\alpha^3$$
$$\log_cd={5\over4} \Rightarrow d=c^{\frac54} =(a-9)^{\frac54} \Rightarrow a-9 \text{ is a perfect 4-th power} \to \alpha^2-9=\beta^4 \to \alpha^2-\beta^4=(\alpha+\beta^2)(\alpha-\beta^2)=9$$
Since $\alpha+\beta^2 \gt 0$ the other factor of $9$ ($\alpha-\beta^2$) must be $\gt0\Rightarrow$
\begin{cases}
\alpha+\beta^2=1, \alpha - \beta^2 =9\\
\alpha+\beta^2=3, \alpha - \beta^2 =3\\
\alpha+\beta^2=9, \alpha - \beta^2 =1
\end{cases}
\begin{cases}
\text{Impossible since } \alpha+\beta^2 \ge \alpha-\beta^2\\
\alpha=3, \beta=0\\
\alpha=5, \beta =2
\end{cases}
\begin{cases}
a=3^2=9, b=3^3=27, c=9-9=0, d= 0^{\frac54}=0 \text{, impossible since } d \gt 0\\
a=5^2=25, b=5^3=125, c= 25-9=16,d=16^{\frac54}=2^5=32
\end{cases}
The only solution is $$a=25\qquad b=125\qquad c=16\qquad d=32$$
$$\Rightarrow b-d=125-32=93$$
