Question

what is another way to write $8^{-4}$?
$8^{-4}=$ ?
$8^{1}=8$
$8^{0}=1$
$8^{-1}=\frac{1}{8}$
$8^{-2}=\frac{1}{8}\cdot\frac{1}{8}$
$8^{-3}=\frac{1}{8}\cdot\frac{1}{8}$

Explanation:

Step1: Recall negative exponent rule

For any non-zero base $a$ and positive integer $n$, $a^{-n} = \frac{1}{a^n}$.

Step2: Apply rule to $8^{-4}$

Substitute $a=8$, $n=4$ into the rule:
$8^{-4} = \frac{1}{8^4}$

Step3: Expand the denominator

$8^4 = 8 \cdot 8 \cdot 8 \cdot 8$, so:
$8^{-4} = \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8}$

Answer:

$\frac{1}{8^4}$ (or $\frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8} \cdot \frac{1}{8}$)