Simplify 2^2x^−1y^−8 9^−1x^0y^−8

Question

. 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: \[ \frac{2^{2/x}}{9y^8} \]