Exercise 5.7 \( \begin{array}{ll}\text { Calculate the following: } & \text { 2. } 5^{3} \times 7^{-2} \times 5^{-3} \times 7^{4} \\ \text { 1. } 2^{2} \times 3 \times 2^{-2} \times 3^{-1} \\ \text { (3) } 2^{3} \times 11^{-4} \times 2^{-2} \times 11^{3} & \text { (4. } 2^{-1} \times 6^{2} \times 3^{-2} \\ \text { (5) } 3^{-3} \times 9 \times 3^{-2} \perp & \text { 8. } 9^{-2} \times 2^{2} \times 3^{-1} \\ \text { 7. } 9^{-1} \times 6^{3} \times 4^{-2} & \text { 10. } 2^{-1} \times 6^{3} \div 3^{2} \\ \text { 9. } 5^{-1} \div 5 & \end{array} \)
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To solve \( 5^{3} \times 7^{-2} \times 5^{-3} \times 7^{4} \), we can combine the powers of 5 and 7. The powers of 5 simplify as \( 5^{3 - 3} = 5^{0} = 1 \). For the powers of 7, we have \( 7^{-2 + 4} = 7^{2} \). Therefore, the expression simplifies to \( 1 \times 7^{2} = 49 \). Next, for \( 2^{2} \times 3 \times 2^{-2} \times 3^{-1} \), we can again combine the powers. The terms with base 2 result in \( 2^{2 - 2} = 2^{0} = 1 \), and for base 3, we get \( 3^{1 - 1} = 3^{0} = 1 \). Thus, the expression simplifies to \( 1 \times 1 = 1 \). If you keep practicing with these exponent rules, you'll cruise through more complex problems like a math wizard!
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