Question

express the following fraction in simplest form, only using positive exponents.
$\frac{12q^{-3}}{(3p^{4}q^{-3})^{-4}}$

Explanation:

Step1: Simplify denominator's negative exponent

Use rule $(x^a)^b=x^{ab}$ and $x^{-a}=\frac{1}{x^a}$:
$$(3p^4q^{-3})^{-4}=3^{-4}p^{-16}q^{12}=\frac{q^{12}}{3^4p^{16}}=\frac{q^{12}}{81p^{16}}$$

Step2: Rewrite division as multiplication by reciprocal

$$\frac{12q^{-3}}{\frac{q^{12}}{81p^{16}}}=12q^{-3} \times \frac{81p^{16}}{q^{12}}$$

Step3: Convert negative exponent to positive

Use rule $x^{-a}=\frac{1}{x^a}$:
$$12 \times \frac{1}{q^3} \times \frac{81p^{16}}{q^{12}}$$

Step4: Multiply constants and combine exponents

Multiply constants: $12 \times 81 = 972$; use rule $\frac{1}{x^a} \times \frac{1}{x^b}=\frac{1}{x^{a+b}}$:
$$\frac{972p^{16}}{q^{3+12}}=\frac{972p^{16}}{q^{15}}$$

Answer:

$\frac{972p^{16}}{q^{15}}$