Question: Order the expressions by choosing <, >, or =. \[ \begin{array}{ccc} 3^{-1} & \square &…

Order the expressions by choosing <, >, or =.

\[ \begin{array}{ccc} 3^{-1} & \square & \left(\frac{1}{3}\right)^{-1} \\ 3^{-2} & \square & \left(\frac{1}{3}\right)^{-1} \\ \left(\frac{1}{3}\right)^{-2} & \square & \left(\frac{1}{3}\right)^{-1} \end{array} \]

\(\square < \square\), \(\square > \square\), \(\square = \square\)

Solution

To solve the problem, we need to simplify each expression and then compare them. 1. First expression: \[ \left(3^{-1}\right) \quad \text{and} \quad \left(\left(\frac{1}{3}\right)^{-1}\right) \] Simplifying \(3^{-1}\): \[ 3^{-1} = \frac{1}{3} \] Simplifying \(\left(\frac{1}{3}\right)^{-1}\): \[ \left(\frac{1}{3}\right)^{-1} = 3 \] Since \(\frac{1}{3} < 3\), the first inequality is: \[ 3^{-1} < \left(\frac{1}{3}\right)^{-1} \] 2. Second expression: \[ \left(3^{-2}\right) \quad \text{and} \quad \left(\left(\frac{1}{3}\right)^{-1}\right) \] Simplifying \(3^{-2}\): \[ 3^{-2} = \frac{1}{9} \] \(\left(\frac{1}{3}\right)^{-1} = 3\) (same as above) Since \(\frac{1}{9} < 3\), the second inequality is: \[ 3^{-2} < \left(\frac{1}{3}\right)^{-1} \] 3. Third expression: \[ \left(\left(\frac{1}{3}\right)^{-2}\right) \quad \text{and} \quad \left(\left(\frac{1}{3}\right)^{-1}\right) \] Simplifying \(\left(\frac{1}{3}\right)^{-2}\): \[ \left(\frac{1}{3}\right)^{-2} = 3^2 = 9 \] \(\left(\frac{1}{3}\right)^{-1} = 3\) (same as above) Since \(9 > 3\), the third inequality is: \[ \left(\frac{1}{3}\right)^{-2} > \left(\frac{1}{3}\right)^{-1} \] To summarize, the answers are: 1. \(3^{-1} < \left(\frac{1}{3}\right)^{-1}\) 2. \(3^{-2} < \left(\frac{1}{3}\right)^{-1}\) 3. \(\left(\frac{1}{3}\right)^{-2} > \left(\frac{1}{3}\right)^{-1}\)