Question

Simplify 2^2x^−1y^−8 9^−1x^0y^−8
. Write your answer using only positive exponents. (6 points)

Answer 1

To simplify the expression \( 2^{2x^{-1}y^{-8}} \cdot 9^{-1}x^{0}y^{-8} \), we can follow these steps:

1. **Rewrite the bases:**
Recall that \( 9 = 3^2 \), so \( 9^{-1} = (3^2)^{-1} = 3^{-2} \).

2. **Rewrite \( x^0 \):**
Since any variable raised to the power of 0 is equal to 1, we have \( x^0 = 1 \).

With that, we can rewrite the expression as:
\[
2^{2x^{-1}y^{-8}} \cdot 3^{-2} \cdot 1 \cdot y^{-8}
\]

3. **Combine the terms with \( y^{-8} \):**
The \( y^{-8} \) part remains the same. So we can express the product as:
\[
2^{2x^{-1}y^{-8}} \cdot 3^{-2} \cdot y^{-8}
\]

4. **Rewrite everything using positive exponents:**
Using the rule that \( a^{-b} = \frac{1}{a^b} \), we can express \( 3^{-2} \) and \( y^{-8} \) with positive exponents:
\[
= \frac{2^{2x^{-1}}}{3^2} \cdot \frac{1}{y^8}
\]

Thus we can collect these terms together:
\[
= \frac{2^{2x^{-1}}}{9y^8}
\]

At this point, we have simplified the expression with everything in positive exponents:
\[
= \frac{2^{2/x}}{9y^8}
\]

So the final result is:
\[
\frac{2^{2/x}}{9y^8}
\]