To find an expanded equivalent expression for \( 7^{15} \) using the Power Rule of Exponents, we will evaluate each of the provided options to see if they equate to \( 7^{15} \).
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Option: \( (7 \cdot 7 \cdot 7 \cdot 7 \cdot 7)^{10} \)
- This can be rewritten as \( 7^5 \) raised to the power of 10, which equals \( 7^{5 \cdot 10} = 7^{50} \).
- This is NOT equal to \( 7^{15} \).
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Option: \( (7^{10})^5 \)
- Using the Power Rule: \( (a^m)^n = a^{m \cdot n} \), this becomes \( 7^{10 \cdot 5} = 7^{50} \).
- This is NOT equal to \( 7^{15} \).
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Option: \( (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \)
- There are 5 groups of \( (7 \cdot 7 \cdot 7) \), each contributing \( 7^3 \). Thus, this expression equals \( (7^3)^5 \).
- Applying the Power Rule: \( 7^{3 \cdot 5} = 7^{15} \).
- This is equal to \( 7^{15} \).
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Option: \( (7^{18})^3 \)
- Using the Power Rule: this becomes \( 7^{18 \cdot 3} = 7^{54} \).
- This is NOT equal to \( 7^{15} \).
Thus, the only expression that simplifies to \( 7^{15} \) is: (7· 7· 7)· (7· 7· 7)· (7· 7· 7)· (7· 7· 7)· (7· 7· 7).