Question

rewrite the expression $4^{-3} \cdot 9^0$ using only positive exponents. rewrite $4^{-3}$ using a positive exponent. $\

$$\begin{align*} 4^{-3} \\cdot 9^0 &= 4^{-3} \\cdot 1 \\\\ &= 4^{-3} \\\\ &=? \\end{align*}$$

$ options: $\frac{1}{4^{-3}}$, $\frac{1}{4^3}$, $-\frac{1}{4^{-3}}$, $-\frac{1}{4^3}$

Explanation:

Step1: Recall the negative exponent rule

The rule for negative exponents is \( a^{-n}=\frac{1}{a^{n}} \) where \( a
eq0 \) and \( n \) is a positive integer.

Step2: Apply the rule to \( 4^{-3} \)

Using the negative exponent rule with \( a = 4 \) and \( n=3 \), we get \( 4^{-3}=\frac{1}{4^{3}} \).

Answer:

\(\frac{1}{4^{3}}\) (corresponding to the option \(\frac{1}{4^{3}}\))