Answer
$a^{8}b^{12}$
Work Step by Step
To solve $(\frac{a^{-2}}{b^{3}})^{-4}$,
RECALL:
(i) The quotients-to-powers rule states that: $(\frac{a}{b})^n=\frac{a^n}{b^n}$
(ii) The power-rule states that $(a^m)^n=a^{mn}$
(iii) The negative-exponent rule states that: $a^{−m} =\frac{1}{a^m}$ and $\frac{1}{a^{-m}} = a^{m}$
Hence, using quotients-to-powers rule and power rule:
$(\frac{a^{-2}}{b^{3}})^{-4}$ = $\frac{(a^{-2})^{-4}}{(b^{3})^{-4}}$
Using the power rule:
$=\frac{a^{-2(-4)}}{b^{3(-4)}}$
$=\frac{a^{8}}{b^{-12}}$
Using negative-exponent rule,
$=\frac{a^{8}}{b^{-12}}= a^{8}b^{12}$
