Answer :
To solve the given expression [tex]\(\left(2^{-4} \cdot z^{-3}\right)^5\)[/tex], we need to apply the properties of exponents. Here's a detailed step-by-step solution:
1. Understand the properties of exponents:
- Power of a product: [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex].
- Power of a power: [tex]\((a^m)^n = a^{m \cdot n}\)[/tex].
2. Apply the power of a product property:
[tex]\[ \left(2^{-4} \cdot z^{-3}\right)^5 = \left(2^{-4}\right)^5 \cdot \left(z^{-3}\right)^5 \][/tex]
3. Apply the power of a power property:
[tex]\[ \left(2^{-4}\right)^5 = 2^{-4 \cdot 5} = 2^{-20} \][/tex]
[tex]\[ \left(z^{-3}\right)^5 = z^{-3 \cdot 5} = z^{-15} \][/tex]
4. Combine the results:
[tex]\[ \left(2^{-4} \cdot z^{-3}\right)^5 = 2^{-20} \cdot z^{-15} \][/tex]
5. Simplify the expression:
Recall that [tex]\(a^{-m} = \frac{1}{a^m}\)[/tex]. Using this property:
[tex]\[ 2^{-20} \cdot z^{-15} = \frac{1}{2^{20}} \cdot \frac{1}{z^{15}} = \frac{1}{2^{20} \cdot z^{15}} \][/tex]
Hence, the equivalent expression is:
[tex]\[ \frac{1}{2^{20} \cdot z^{15}} \][/tex]
So, the correct answer is:
(B) [tex]\(\frac{1}{2^{20} \cdot z^{15}}\)[/tex].
1. Understand the properties of exponents:
- Power of a product: [tex]\((a \cdot b)^n = a^n \cdot b^n\)[/tex].
- Power of a power: [tex]\((a^m)^n = a^{m \cdot n}\)[/tex].
2. Apply the power of a product property:
[tex]\[ \left(2^{-4} \cdot z^{-3}\right)^5 = \left(2^{-4}\right)^5 \cdot \left(z^{-3}\right)^5 \][/tex]
3. Apply the power of a power property:
[tex]\[ \left(2^{-4}\right)^5 = 2^{-4 \cdot 5} = 2^{-20} \][/tex]
[tex]\[ \left(z^{-3}\right)^5 = z^{-3 \cdot 5} = z^{-15} \][/tex]
4. Combine the results:
[tex]\[ \left(2^{-4} \cdot z^{-3}\right)^5 = 2^{-20} \cdot z^{-15} \][/tex]
5. Simplify the expression:
Recall that [tex]\(a^{-m} = \frac{1}{a^m}\)[/tex]. Using this property:
[tex]\[ 2^{-20} \cdot z^{-15} = \frac{1}{2^{20}} \cdot \frac{1}{z^{15}} = \frac{1}{2^{20} \cdot z^{15}} \][/tex]
Hence, the equivalent expression is:
[tex]\[ \frac{1}{2^{20} \cdot z^{15}} \][/tex]
So, the correct answer is:
(B) [tex]\(\frac{1}{2^{20} \cdot z^{15}}\)[/tex].
