Question

simplify. express your answer using exponents.
$(j^{7}k^{-2})^{-9}$

Explanation:

Step1: Apply power of a product rule

$(ab)^n = a^n b^n$, so:
$(j^7 k^{-2})^{-9} = (j^7)^{-9} \cdot (k^{-2})^{-9}$

Step2: Apply power of a power rule

$(a^m)^n = a^{m \cdot n}$, so:
$(j^7)^{-9} = j^{7 \times (-9)}$, $(k^{-2})^{-9} = k^{(-2) \times (-9)}$

Step3: Calculate exponents

$j^{7 \times (-9)} = j^{-63}$, $k^{(-2) \times (-9)} = k^{18}$

Step4: Rewrite positive exponents (optional, standard form)

$j^{-63} = \frac{1}{j^{63}}$, so combine terms:
$\frac{k^{18}}{j^{63}}$ or $j^{-63}k^{18}$

Answer:

$\frac{k^{18}}{j^{63}}$ (or $j^{-63}k^{18}$)