Intermediate Algebra, Chapter 5, 5.1, Section 5.1, Problem 55

Evaluate the expression $\dfrac{2^{-2}}{3^{-3}}$.
Remove the negative exponent by rewriting $2^{-2}$ as $\dfrac{1}{2^2}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{2^2}\cdot\dfrac{1}{3^{-3}}$
Squaring a number is the same as multiplying the number by itself $2\cdot2$. In this case, $2$ squared is $4$.
$\dfrac{1}{4}\cdot\dfrac{1}{3^{-3}}$
Remove the negative exponent by rewriting $3^{-3}$ as $\dfrac{1}{3^3}$. A negative exponent follows the rule of $a^{-n} = \dfrac{1}{a^n}$
$\dfrac{1}{4}\cdot\dfrac{1}{\dfrac{1}{3^{3}}}$
Remove the parentheses from the denominator.
$\dfrac{1}{4}\cdot\dfrac{1}{\dfrac{1}{27}}$
Simplify the multi-level fraction by multiplying by the reciprocal of the denominator.
$\dfrac{1}{4}\cdot 27$
Multiply $\dfrac{1}{4}$ by $27$ to get $\dfrac{27}{4}$ .
$\dfrac{27}{4}$