What value of n makes the equation true? (2x^9 y^n)(4x^2 y^10) = 8x^11 y^20

1. Simplify the left side of the equation:
We are given the equation $(2x^9 y^n)(4x^2 y^10) = 8x^{11} y^{20}$. To solve for $n$, we first simplify the left side by multiplying the coefficients and adding the exponents of the like bases.

Multiply the coefficients: $2 \times 4 = 8$.
Add the exponents of $x$: $x^9 \times x^2 = x^{9+2} = x^{11}$.
Add the exponents of $y$: $y^n \times y^{10} = y^{n+10}$.

So, the simplified left side is $8x^{11} y^{n+10}$.

2. Equate the simplified left side with the right side:
Now we have the equation $8x^{11} y^{n+10} = 8x^{11} y^{20}$.

3. Solve for n by comparing the exponents of y:
For the equation to be true, the exponents of the corresponding variables on both sides must be equal. We can see that the coefficients (8) and the powers of $x$ ($x^{11}$) are already equal.

We need to equate the exponents of $y$: $n+10 = 20$.

Subtract 10 from both sides of the equation:
$n+10 - 10 = 20 - 10$
$n = 10$

4. Verify the solution:
Substitute $n=10$ back into the original equation:
$(2x^9 y^{10})(4x^2 y^{10}) = 8x^{9+2} y^{10+10} = 8x^{11} y^{20}$.
This matches the right side of the given equation, so our solution is correct.