If $a-c = 9$ then find the value of $b-d$.

Question

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?

Answer 1

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 $p=5$.

$b=p^3=125$; $d=q^5=32$. $b-d=125-32=93$.

Answer 2

$$a=b^\frac{2}{3}$$

$$c=d^\frac{4}{5}$$

$$a-c=b^\frac{2}{3}-d^\frac{4}{5}$$

Clearly $$b=5^3$$ and

$$d=2^5$$

satisfies the equation.

Answer 3

Trial and error may be the best way to go about it. Regardless, the answer isn't too hard to find, especially if you're thinking in terms of perfect squares. $25 - 16 = 9$, and thus:

$$b - d = 93$$