As Exponents (Part 2)

In $2^{x-8}5^{y-8}=32$ , the product only has $2$ as its prime factor. Because of this, $5^{y-8}$ should be equal to $1$ .

$\large \begin{aligned} 5^{y-8} &= 1 \\ 5^{8-8} &= 1 \\ 5^{0} &= 1 \\ \implies y &= 8 \end{aligned}$

This leaves us with $2^{x-8}=32$ . Since $32=2^{5}$ :

$\large \begin{aligned} 2^{x-8} &= 32 \\ 2^{x-8} &= 2^{5} \\ x-8 &= 5 \\ x &= 13 \end{aligned}$

Getting the values of $x$ and $y$ , we can solve $\frac{x}{y}$ which is $\frac{13}{8}$ or $\boxed{1.625}$ .

The same method can be done to the second product in the problem, with $2^{y-8}$ equalling to $1$ since $3125$ has $5$ as its only prime factor.

From $2^{x-8}5^{y-8} = 32 = 2^5\times 5^0$ . Since $x$ and $y$ are positive integers, this implies that $x-8=5 \implies x = 13$ and $y-8 = 0 \implies y = 8$ . Note that $x=13$ and $y=8$ also satisfy the second equation $2^{8-8} 5^{13-8} = 5^5 = 3125$ . Therefore, $\dfrac xy = \dfrac {13}8 = \boxed{1.625}$ .