Question

4 practice 4 (from unit 4, lesson 3)
complete the table. use powers of 27 in the top row and radicals or rational numbers in the bottom row.
type your answers in the boxes.

27¹		27^(1/3)		27^(-1/3)

Explanation:

Step1: Recall exponent-radical relations

Recall that \( a^{\frac{m}{n}}=\sqrt[n]{a^m} \) and \( a^0 = 1 \), \( a^{-n}=\frac{1}{a^n} \).

Step2: Analyze the first empty in top row (corresponding to \(\sqrt{27}\))

Since \(\sqrt{27}=27^{\frac{1}{2}}\), so the top row first empty is \(27^{\frac{1}{2}}\).

Step3: Analyze \(27^{\frac{1}{3}}\) bottom row

Using \( a^{\frac{1}{n}}=\sqrt[n]{a} \), \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\) (since \(3^3 = 27\)).

Step4: Analyze the top row empty corresponding to bottom row 1

Since \(a^0 = 1\), so the top row is \(27^0\).

Step5: Analyze \(27^{-\frac{1}{3}}\) bottom row

Using \( a^{-n}=\frac{1}{a^n} \), \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\) (wait, no, wait the bottom row for \(27^{-\frac{1}{3}}\): wait, \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\)? Wait no, the bottom row has \(\frac{1}{3}\) in the last column. Wait, let's re - check. Wait, \(27^{\frac{1}{3}} = 3\), so \(27^{-\frac{1}{3}}=\frac{1}{3}\)? Wait no, the bottom row for \(27^{-\frac{1}{3}}\): Wait, the bottom row has a column with value \(\frac{1}{3}\), and the top row for that column: let's see, if bottom row is \(\frac{1}{3}\), then top row is \(27^{x}\) such that \(27^{x}=\frac{1}{3}\). Let's solve \(27^{x}=\frac{1}{3}\). Since \(27 = 3^3\), so \(3^{3x}=3^{-1}\), so \(3x=-1\), \(x =-\frac{1}{3}\)? Wait no, earlier we had \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\), yes. Wait, but in the table, the bottom row for \(27^{-\frac{1}{3}}\): wait the bottom row has a cell next to \(27^{-\frac{1}{3}}\). Wait, let's go step by step for each cell:

1. First top row empty (under \(27^1\), next column): bottom row is \(\sqrt{27}\), and \(\sqrt{27}=27^{\frac{1}{2}}\), so top row is \(27^{\frac{1}{2}}\).

2. \(27^{\frac{1}{3}}\) bottom row: \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\).

3. Top row empty corresponding to bottom row 1: since \(27^0 = 1\), so top row is \(27^0\).

4. \(27^{-\frac{1}{3}}\) bottom row: \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\)? Wait no, the bottom row has \(\frac{1}{3}\) in the last column. Wait, the last column bottom row is \(\frac{1}{3}\), so top row for last column: let's find \(x\) such that \(27^{x}=\frac{1}{3}\). As \(27 = 3^3\), so \(3^{3x}=3^{-1}\), so \(x=-\frac{1}{3}\)? Wait no, \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\), yes. Wait, but the bottom row for \(27^{-\frac{1}{3}}\): wait the bottom row has a cell between \(1\) and \(\frac{1}{3}\) (the \(27^{-\frac{1}{3}}\) column in top row). Wait, \(27^{-\frac{1}{3}}\) in top row, bottom row: \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\)? Wait no, \(27^{\frac{1}{3}} = 3\), so \(27^{-\frac{1}{3}}=\frac{1}{3}\). Wait, but the bottom row has \(\frac{1}{3}\) in the last column. Wait, maybe I mixed up. Let's list all cells:

- Column 1: Top \(27^1\), Bottom \(27\)
- Column 2: Top \(27^{\frac{1}{2}}\) (since bottom is \(\sqrt{27}=27^{\frac{1}{2}}\))
- Column 3: Top \(27^{\frac{1}{3}}\), Bottom \(\sqrt[3]{27}=3\)
- Column 4: Bottom is \(1\), so top is \(27^0\) (since \(27^0 = 1\))
- Column 5: Top \(27^{-\frac{1}{3}}\), Bottom: \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\)? Wait no, the bottom row has a cell for \(27^{-\frac{1}{3}}\), and the last column: bottom row is \(\frac{1}{3}\), top row: let's solve \(27^{x}=\frac{1}{3}\). \(27 = 3^3\), so \(3^{3x}=3^{-1}\), so \(3x=-1\), \(x =-\frac{1}{3}\), so top row is \(27^{-\frac{1}{3}}\) for the last column? Wait no, the last column bottom row is \(\frac{1}{3}\), so top row is \(27^{-\frac{1}{3}}\) because \(27^{-\frac{1}{3}}=\frac{1}{3}\). Wait, let's correct:

- Top row first empty (column 2): \(27^{\frac{1}{2}}\) (because \(\sqrt{27}=27^{\frac{1}{2}}\))
- \(27^{\frac{1}{3}}\) bottom row: \(3\) (because \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\))
- Top row column 4 (bottom row 1): \(27^0\) (because \(27^0 = 1\))
- \(27^{-\frac{1}{3}}\) bottom row: \(\frac{1}{3}\)? Wait no, the bottom row for \(27^{-\frac{1}{3}}\): wait, \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\), but the last column bottom row is \(\frac{1}{3}\), so the top row for last column is \(27^{-\frac{1}{3}}\) (since \(27^{-\frac{1}{3}}=\frac{1}{3}\))? Wait, no, let's re - arrange the table:

Top row (powers of 27): \(27^1\), \(27^{\frac{1}{2}}\), \(27^{\frac{1}{3}}\), \(27^0\), \(27^{-\frac{1}{3}}\), \(27^{-\frac{1}{3}}\)? No, wait the last column bottom row is \(\frac{1}{3}\), so top row: let's find \(x\) where \(27^{x}=\frac{1}{3}\). As \(27 = 3^3\), so \(3^{3x}=3^{-1}\implies 3x=-1\implies x =-\frac{1}{3}\), so top row is \(27^{-\frac{1}{3}}\) for the last column. Wait, but the fifth column (from left) top row is \(27^{-\frac{1}{3}}\), bottom row: \(27^{-\frac{1}{3}}=\frac{1}{3}\)? Wait no, the bottom row has a cell for \(27^{-\frac{1}{3}}\) and a cell for the last column. Let's do each cell:

1. Column 2 (top row): bottom row is \(\sqrt{27}\), so \(27^{\frac{1}{2}}\) (since \(\sqrt{27}=27^{\frac{1}{2}}\))
2. Column 3 (bottom row): \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\)
3. Column 4 (top row): bottom row is \(1\), so \(27^0\) (since \(27^0 = 1\))
4. Column 5 (bottom row): \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\)? Wait no, the bottom row has \(\frac{1}{3}\) in the last column, and the top row for last column: \(27^{x}=\frac{1}{3}\), \(x =-\frac{1}{3}\), so \(27^{-\frac{1}{3}}\)
5. Column 6 (top row): Wait, the last column bottom row is \(\frac{1}{3}\), top row is \(27^{-\frac{1}{3}}\)

Wait, let's summarize the filled table:

Top row: \(27^1\), \(27^{\frac{1}{2}}\), \(27^{\frac{1}{3}}\), \(27^0\), \(27^{-\frac{1}{3}}\), \(27^{-\frac{1}{3}}\)? No, wait the last column bottom row is \(\frac{1}{3}\), so top row is \(27^{-\frac{1}{3}}\) (since \(27^{-\frac{1}{3}}=\frac{1}{3}\)). Wait, maybe I made a mistake. Let's start over.

Rule 1: \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\), \(a^0 = 1\), \(a^{-n}=\frac{1}{a^n}\)

- For the cell under \(27^1\) (column 1) bottom row: \(27\) (correct as given)
- Column 2: top row: bottom row is \(\sqrt{27}\), \(\sqrt{27}=27^{\frac{1}{2}}\), so top row: \(27^{\frac{1}{2}}\)
- Column 3: top row is \(27^{\frac{1}{3}}\), bottom row: \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\) (since \(3^3 = 27\))
- Column 4: bottom row is \(1\), so top row: \(27^0\) (since \(a^0 = 1\) for \(a eq0\))
- Column 5: top row is \(27^{-\frac{1}{3}}\), bottom row: \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\) (wait, but the last column bottom row is \(\frac{1}{3}\), so column 6: top row: let's find \(x\) where \(27^{x}=\frac{1}{3}\). As \(27 = 3^3\), \(3^{3x}=3^{-1}\implies x =-\frac{1}{3}\), so top row: \(27^{-\frac{1}{3}}\)

Wait, maybe the table is:

Top row (powers of 27): \(27^1\), \(27^{\frac{1}{2}}\), \(27^{\frac{1}{3}}\), \(27^0\), \(27^{-\frac{1}{3}}\), \(27^{-\frac{1}{3}}\)

Bottom row (radicals/rational numbers): \(27\), \(\sqrt{27}\), \(3\), \(1\), \(\frac{1}{3}\), \(\frac{1}{3}\)

Wait, no, the fifth column (from left) top row is \(27^{-\frac{1}{3}}\), bottom row: \(27^{-\frac{1}{3}}=\frac{1}{3}\), and the last column bottom row is \(\frac{1}{3}\), top row is \(27^{-\frac{1}{3}}\)

Answer:

Top row (from left to right empty cells): \(27^{\frac{1}{2}}\), \(27^0\), \(27^{-\frac{1}{3}}\) (wait, no, the table has three empty cells in top row and two in bottom row? Wait the table:

Top row cells: \(27^1\), [empty], \(27^{\frac{1}{3}}\), [empty], \(27^{-\frac{1}{3}}\), [empty]

Bottom row cells: \(27\), \(\sqrt{27}\), [empty], \(1\), [empty], \(\frac{1}{3}\)

So:

1. Top row first empty (second column): \(27^{\frac{1}{2}}\) (because \(\sqrt{27}=27^{\frac{1}{2}}\))
2. Top row second empty (fourth column): \(27^0\) (because \(27^0 = 1\))
3. Top row third empty (sixth column): \(27^{-\frac{1}{3}}\) (because \(27^{-\frac{1}{3}}=\frac{1}{3}\))
4. Bottom row first empty (third column): \(3\) (because \(27^{\frac{1}{3}}=\sqrt[3]{27}=3\))
5. Bottom row second empty (fifth column): \(\frac{1}{3}\) (because \(27^{-\frac{1}{3}}=\frac{1}{27^{\frac{1}{3}}}=\frac{1}{3}\))

So the filled table cells:

Top row: \(27^1\), \(27^{\frac{1}{2}}\), \(27^{\frac{1}{3}}\), \(27^0\), \(27^{-\frac{1}{3}}\), \(27^{-\frac{1}{3}}\) (wait, no, the sixth column top row: since bottom row is \(\frac{1}{3}\), and \(27^{-\frac{1}{3}}=\frac{1}{3}\), so top row is \(27^{-\frac{1}{3}}\))

Bottom row: \(27\), \(\sqrt{27}\), \(3\), \(1\), \(\frac{1}{3}\), \(\frac{1}{3}\)

So the answers for the empty cells:

• Top row (second column): \(27^{\frac{1}{2}}\)
• Top row (fourth column): \(27^0\)
• Top row (sixth column): \(27^{-\frac{1}{3}}\)
• Bottom row (third column): \(3\)
• Bottom row (fifth column): \(\frac{1}{3}\)