To simplify the expression \((7^{-3})^7 \cdot 7^{20} / 6^{-2}\) using the properties of integer exponents and ensuring all exponents are positive, we will follow these steps:
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Apply the Power of a Power Property: \((a^m)^n = a^{m \cdot n}\). So, we calculate: \[ (7^{-3})^7 = 7^{-21} \]
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Substituting back into the expression: Now the expression becomes: \[ 7^{-21} \cdot 7^{20} / 6^{-2} \]
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Combine the terms with the same base: We can add the exponents for the \(7\) terms: \[ 7^{-21} \cdot 7^{20} = 7^{-21 + 20} = 7^{-1} \]
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Rewrite \(6^{-2}\): Remember that \(a^{-m} = \frac{1}{a^m}\): \[ 6^{-2} = \frac{1}{6^2} \] Therefore: \[ \frac{1}{6^{-2}} = 6^2 = 36 \]
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Combine the entire expression: Now the expression becomes: \[ \frac{7^{-1}}{6^{-2}} = 7^{-1} \cdot 6^2 = \frac{6^2}{7} \]
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Calculate \(6^2\): \[ 6^2 = 36 \]
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Final expression: Thus, we have: \[ \frac{6^2}{7} = \frac{36}{7} \]
Hence, the simplified expression is: \[ \frac{36}{7} \]
This is the final answer as an improper fraction.
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