To simplify the given expression \( (7^{-3})^7 \times \frac{7^{20}}{6^{-2}} \) using the properties of integer exponents and to ensure we have only positive exponents, we can follow these steps:
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Apply the Power of a Power Property: \[ (a^m)^n = a^{m \cdot n} \] Therefore, \[ (7^{-3})^7 = 7^{-3 \cdot 7} = 7^{-21} \]
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Rewrite the expression: Now the expression becomes: \[ 7^{-21} \times \frac{7^{20}}{6^{-2}} \]
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Rewriting the division: Recall that \[ \frac{a^m}{b^n} = a^m \cdot b^{-n} \] Thus, the fraction \(\frac{7^{20}}{6^{-2}}\) can be rewritten as: \[ 7^{20} \cdot 6^{2} \] So now our expression is: \[ 7^{-21} \times 7^{20} \times 6^{2} \]
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Combine the like bases: Use the property \( a^m \cdot a^n = a^{m+n} \): \[ 7^{-21} \times 7^{20} = 7^{-21 + 20} = 7^{-1} \]
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Substituting back into the expression: Therefore, we now have: \[ 7^{-1} \times 6^2 \]
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Convert to positive exponents: Rewrite \(7^{-1}\) as \(\frac{1}{7}\): \[ \frac{1}{7} \times 6^2 = \frac{1}{7} \times 36 = \frac{36}{7} \]
Thus, the final result, expressed as an improper fraction, is: \[ \boxed{\frac{36}{7}} \]
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