Subjects algebra

Equivalent Expressions 18022B

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Question: Select all the pairs of equivalent expressions. $4^{-3} \cdot 4$ and $16^{-2}$ $4^{-3} \cdot 5^{2}$ and $20^{-1}$ $\frac{3^{5}}{3^{8}}$ and $3^{-3}$ $(5^{2})^{-6}$ and $5^{-4}$ $\left(\frac{3}{4}\right)^{-2}$ and $\left(\frac{4}{3}\right)^{2}$
1. **State the problem:** We need to determine which pairs of expressions are equivalent. 2. **Recall exponent rules:** - $a^{m} \cdot a^{n} = a^{m+n}$ - $\frac{a^{m}}{a^{n}} = a^{m-n}$ - $(a^{m})^{n} = a^{m \cdot n}$ - $a^{-m} = \frac{1}{a^{m}}$ - $\left(\frac{a}{b}\right)^{-m} = \left(\frac{b}{a}\right)^{m}$ --- 3. **Check each pair:** **Pair 1:** $4^{-3} \cdot 4$ and $16^{-2}$ - Simplify $4^{-3} \cdot 4 = 4^{-3} \cdot 4^{1} = 4^{-3+1} = 4^{-2}$ - Simplify $16^{-2} = (4^{2})^{-2} = 4^{2 \cdot (-2)} = 4^{-4}$ - Since $4^{-2} \neq 4^{-4}$, these are **not equivalent**. **Pair 2:** $4^{-3} \cdot 5^{2}$ and $20^{-1}$ - Simplify $4^{-3} \cdot 5^{2}$ stays as is. - Simplify $20^{-1} = \frac{1}{20} = \frac{1}{4 \cdot 5}$ - $4^{-3} = \frac{1}{4^{3}} = \frac{1}{64}$ and $5^{2} = 25$, so $4^{-3} \cdot 5^{2} = \frac{25}{64}$ - $20^{-1} = \frac{1}{20} = 0.05$, $\frac{25}{64} \approx 0.390625$ - Not equal, so **not equivalent**. **Pair 3:** $\frac{3^{5}}{3^{8}}$ and $3^{-3}$ - Simplify $\frac{3^{5}}{3^{8}} = 3^{5-8} = 3^{-3}$ - Both sides equal $3^{-3}$, so **equivalent**. **Pair 4:** $(5^{2})^{-6}$ and $5^{-4}$ - Simplify $(5^{2})^{-6} = 5^{2 \cdot (-6)} = 5^{-12}$ - Compare with $5^{-4}$ - Since $5^{-12} \neq 5^{-4}$, **not equivalent**. **Pair 5:** $\left(\frac{3}{4}\right)^{-2}$ and $\left(\frac{4}{3}\right)^{2}$ - Simplify $\left(\frac{3}{4}\right)^{-2} = \left(\frac{4}{3}\right)^{2}$ by the negative exponent rule - Both sides are equal, so **equivalent**. --- **Final answer:** The equivalent pairs are: - $\frac{3^{5}}{3^{8}}$ and $3^{-3}$ - $\left(\frac{3}{4}\right)^{-2}$ and $\left(\frac{4}{3}\right)^{2}$