Questions: Order the expressions by choosing <, >, or =. 6^-1 □ 6^-2 (1/6)^-1 □ 6^-2 (1/6)^-1 □ (1/6)^-2

Order the expressions by choosing <, >, or =.
6^-1 □ 6^-2
(1/6)^-1 □ 6^-2
(1/6)^-1 □ (1/6)^-2

Transcript text: Order the expressions by choosing $\langle$,$\rangle , or =$. $6^{-1} \square 6^{-2}$ $\left(\frac{1}{6}\right)^{-1} \square 6^{-2}$ $\left(\frac{1}{6}\right)^{-1} \square\left(\frac{1}{6}\right)^{-2}$ Explanation Check

Solution

Solution Steps

Step 1: Convert to Common Bases

Given expressions are $a^{-n} = 6^{-1}$ and $\left(\frac{1}{a}\right)^{-m} = 6^{2}$. We simplify the second expression by recognizing that taking a negative exponent inverts the base, thus $\left(\frac{{1}}{{a}}\right)^{{-m}}$ becomes $a^m$.

Step 2: Compare Exponents

Since $-n < m$, it implies that $a^{-n} > a^m$ because smaller (more negative) exponents mean smaller values when $a > 1$.

Final Answer: $a^{-n} > a^m$

Step 1: Convert to Common Bases

Given expressions are $a^{-n} = 6^{-1}$ and $\left(\frac{1}{a}\right)^{-m} = 6^{2}$. We simplify the second expression by recognizing that taking a negative exponent inverts the base, thus $\left(\frac{{1}}{{a}}\right)^{{-m}}$ becomes $a^m$.

Step 2: Compare Exponents

Since $-n < m$, it implies that $a^{-n} > a^m$ because smaller (more negative) exponents mean smaller values when $a > 1$.

Final Answer: $a^{-n} > a^m$

Step 1: Convert to Common Bases

Given expressions are $a^{-n} = 6^{-1}$ and $\left(\frac{1}{a}\right)^{-m} = 6^{2}$. We simplify the second expression by recognizing that taking a negative exponent inverts the base, thus $\left(\frac{{1}}{{a}}\right)^{{-m}}$ becomes $a^m$.

Step 2: Compare Exponents

Since $-n < m$, it implies that $a^{-n} > a^m$ because smaller (more negative) exponents mean smaller values when $a > 1$.

Final Answer: $a^{-n} > a^m$

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