Question

lesson 3 practice problems

1. complete the table. use powers of 64 in the top row and radicals or rational numbers

in the bottom row.
\

$$\begin{tabular}{|c|c|c|c|c|c|} \\hline $64^1$ & $64^{\\frac{1}{3}}$ & & $64^0$ & & $64^{-1}$ \\\\ \\hline 64 & & 4 & & $\\frac{1}{8}$ & \\\\ \\hline \\end{tabular}$$

1. suppose that a friend missed class and never learned what $25^{\frac{1}{2}}$ means.

a. use exponent rules your friend would already know to calculate $25^{\frac{1}{2}} \cdot 25^{\frac{1}{2}}$.
b. explain why this means that $25^{\frac{1}{2}} = 5$.

1. which expression is equivalent to $16^{\frac{1}{2}}$?

a. $\frac{1}{4}$
b. 4
c. 8
d. 16.5

Explanation:

Problem 1: Completing the Table

Step 1: Recall Exponent Rules

Recall that \( a^{\frac{m}{n}}=\sqrt[n]{a^m} \) and \( a^0 = 1 \), \( a^{-n}=\frac{1}{a^n} \). Also, \( 64 = 2^6 \), \( 4=2^2=64^{\frac{1}{3}} \) (since \( 64^{\frac{1}{3}}=\sqrt[3]{64} = 4 \)), \( 64^0 = 1 \), \( 64^{-1}=\frac{1}{64} \), and \( 64^{\frac{1}{2}}=\sqrt{64}=8 \), \( 64^{\frac{-1}{3}}=\frac{1}{4} \) (wait, no, the bottom row has \( \frac{1}{8} \), so let's find the exponent for \( \frac{1}{8} \). Since \( \frac{1}{8}=\frac{1}{64^{\frac{1}{2}}}? \) Wait, \( 64^{\frac{-1}{2}}=\frac{1}{8} \) (because \( 64^{\frac{1}{2}} = 8 \), so \( 64^{\frac{-1}{2}}=\frac{1}{8} \)). Let's fill each cell:

• For \( 64^{\frac{1}{2}} \): \( 64^{\frac{1}{2}}=\sqrt{64}=8 \)
• For the cell with 4: We know \( 64^x = 4 \). Since \( 64 = 4^3 \), so \( (4^3)^x=4 \implies 4^{3x}=4^1 \implies 3x = 1 \implies x=\frac{1}{3} \), so the top row is \( 64^{\frac{1}{3}} \)
• For \( 64^0 \): The bottom row is \( 1 \) (since \( a^0 = 1 \))
• For the cell with \( \frac{1}{8} \): We know \( 64^x=\frac{1}{8} \). Since \( 8 = 64^{\frac{1}{2}} \), so \( \frac{1}{8}=64^{-\frac{1}{2}} \), so the top row is \( 64^{-\frac{1}{2}} \)
• For \( 64^{-1} \): \( 64^{-1}=\frac{1}{64} \)

Step 2: Fill the Table

\( 64^1 \)	\( 64^{\frac{1}{2}} \)	\( 64^{\frac{1}{3}} \)	\( 64^0 \)	\( 64^{-\frac{1}{2}} \)	\( 64^{-1} \)

Problem 2a: Calculating \( 25^{\frac{1}{2}} \cdot 25^{\frac{1}{2}} \)

Step 1: Use Exponent Rule \( a^m \cdot a^n = a^{m + n} \)

We know that for any base \( a \) and exponents \( m, n \), \( a^m \cdot a^n=a^{m + n} \). Here, \( a = 25 \), \( m=\frac{1}{2} \), \( n=\frac{1}{2} \). So:

\( 25^{\frac{1}{2}} \cdot 25^{\frac{1}{2}}=25^{\frac{1}{2}+\frac{1}{2}}=25^1 = 25 \)

Step 2: Simplify

So the product is 25.

Problem 2b: Explaining \( 25^{\frac{1}{2}} = 5 \)

Let \( x = 25^{\frac{1}{2}} \). Then, from part (a), \( x \cdot x=25 \), which means \( x^2 = 25 \). The positive square root of 25 is 5 (since we are dealing with real numbers and the principal root), so \( x = 5 \). Thus, \( 25^{\frac{1}{2}} = 5 \).

Problem 3: Equivalent to \( 16^{\frac{1}{2}} \)

Answer:

(Problem 3):
B. 4