To solve for $xy$, we first need to determine the values of $x$ and $y$ individually.

Step 1: Solve $2^x = 32$

We know that $32 = 2^5$. So, by comparing the exponents: $2^x = 2^5 \quad \Rightarrow \quad x = 5$

Step 2: Solve $3^y = 81$

We know that $81 = 3^4$. So, by comparing the exponents: $3^y = 3^4 \quad \Rightarrow \quad y = 4$

Step 3: Calculate $xy$

Now that we have $x = 5$ and $y = 4$, we can calculate $xy$: $xy = 5 \times 4 = 20$

Thus, the value of $xy$ is $\boxed{20}$.

Do you have any further questions or need more details?

Here are 5 related questions you might want to explore:

1. What is the solution to $2^{x+1} = 64$?
2. What is the value of $5^z = 625$?
3. What if $2^x = 64$ and $5^y = 25$? What is $xy$?
4. How do you solve logarithmic equations like $\log_2(32) = x$?
5. What is the relationship between exponential and logarithmic functions?

Tip: To solve exponential equations like $a^x = b$, try expressing $b$ as a power of $a$ if possible, and then compare exponents.