To simplify \(7^{15}\) using the Power Rule of Exponents, it means that we are looking to express it as a product of bases raised to specific powers. The Power Rule states that \( (a^m)^n = a^{m \cdot n} \).
Among the options given, we want to find an expression that, when simplified, would return \(7^{15}\).
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For \((7^{18})^3\): \[ (7^{18})^3 = 7^{18 \cdot 3} = 7^{54} \quad \text{(not equivalent)} \]
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For \((7^{10})^5\): \[ (7^{10})^5 = 7^{10 \cdot 5} = 7^{50} \quad \text{(not equivalent)} \]
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For \((7 \cdot 7 \cdot 7 \cdot 7 \cdot 7)^{10}\): \[ (7 \cdot 7 \cdot 7 \cdot 7 \cdot 7) = 7^5 \quad \text{(so now we raise it to the 10th power)} \] \[ (7^5)^{10} = 7^{5 \cdot 10} = 7^{50} \quad \text{(not equivalent)} \]
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For \((7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7) \cdot (7 \cdot 7 \cdot 7)\): \[ (7 \cdot 7 \cdot 7) = 7^3 \quad \text{(and there are 5 groups of \(7^3\))} \] \[ (7^3)^5 = 7^{3 \cdot 5} = 7^{15} \quad \text{(this is equivalent!)} \]
Thus, the correct expanded equivalent expression for \( 7^{15} \) is:
(7⋅7⋅7)⋅(7⋅7⋅7)⋅(7⋅7⋅7)⋅(7⋅7⋅7)⋅(7⋅7⋅7).