We find the reciprocal of the base $({\color{red}3}),$ and then raise that reciprocal to the same power but with the opposite sign $({\color{blue}+2}),$ as follows: \begin{align} {\color{red}3}^{-\color{blue}2} & = \left( \dfrac{1}{\color{red}3} \right)^{\color{blue}2} \\[5pt] & = \left( \dfrac{1}{3} \right) \times \left( \dfrac{1}{3} \right) \\[5pt] & = \left( \dfrac{1 \times 1}{3 \times 3} \right) \\[5pt] & = \dfrac{1}{9} \end{align}

We find the reciprocal of the base $({\color{red}{-3}}),$ and then raise that reciprocal to the same power but with the opposite sign $({\color{blue}{+3}}),$ as follows:

\begin{align} ({\color{red}{-3}})^{\color{blue}-3} &=\left(\dfrac{1}{\color{red}-3}\right)^{\color{blue}3}\\[5pt] &=\left(-\dfrac{1}{3}\right)^{3}\\[5pt] &=\left(-\dfrac{1}{3}\right) \times \left(-\dfrac{1}{3}\right) \times \left(-\dfrac{1}{3}\right) \\[5pt] &=\dfrac{(-1) \times (-1)\times (-1)}{3 \times 3\times 3}\\[5pt] &=-\dfrac{1}{27} \end{align}

We find the reciprocal of the base $({\color{red}2}),$ and then raise that reciprocal to the same power but with the opposite sign $({\color{blue}+4}),$ as follows: \begin{align} {\color{red}2}^{-{\color{blue}4}} & = \left( \dfrac{1}{\color{red}2} \right)^{\color{blue}4} \\[5pt] & = \left( \dfrac{1}{2} \right) \times \left( \dfrac{1}{2} \right) \times \left( \dfrac{1}{2} \right) \times \left( \dfrac{1}{2} \right)\\[5pt] &= \left( \dfrac{1 \times 1 \times 1 \times 1}{2 \times 2 \times 2 \times 2} \right) \\[5pt] & = \dfrac{1}{16} \end{align}

We find the reciprocal of the base $\left(\dfrac 1 9\right),$ and then raise that reciprocal to the same power but with the opposite sign $(+{\color{blue}2}),$ as follows:

\begin{align} \left(\dfrac{1}{9}\right)^{-\color{blue}2} &=(9)^{\color{blue}2} \\[5pt] &=9 \cdot 9 \\[5pt] &= 81 \end{align}

We find the reciprocal of the base $\left(\dfrac{1}{2}\right)$, and then raise that reciprocal to the same power but with the opposite sign $(+{\color{blue}{3}})$, as follows:

\begin{align*} \left(\dfrac 1 2\right)^{-{\color{blue}{3}}} &= (2)^{\color{blue}3}\\[5pt] &= 2\cdot 2 \cdot 2\\[5pt] &= 8 \end{align*}

We find the reciprocal of the base $\left(\dfrac{1}{2}\right)$, and then raise that reciprocal to the same power but with the opposite sign $(+{\color{blue}4})$, as follows:

\begin{align*} \left(\dfrac 1 2\right)^{-\color{blue}4} &= (2)^{\color{blue}4} \\[5pt] &= 2 \cdot 2 \cdot 2 \cdot 2 \\[5pt] &= 16 \end{align*}

First, we write $\dfrac{1}{25}$ as a base raised to a power:

\begin{align*} \dfrac {1} {25} &= \dfrac{1 \times 1}{5 \times 5}\\[5pt] &= \dfrac{1}{5} \times \dfrac{1}{5}\\[5pt] &= \left( \dfrac{1}{5} \right)^2 \end{align*}

Now, we find the reciprocal of the base $\left(\dfrac{1}{5}\right),$ and then raise that reciprocal to the same power but with the opposite sign $(-2),$ as follows: \[ \left( \dfrac{1}{5} \right)^2 = 5^{-2} \]

First, we write $\dfrac{9}{4}$ as a base raised to a power:

\begin{align*} \dfrac {9} {4} &= \dfrac{3 \times 3}{2 \times 2}\\[5pt] &= \dfrac{3}{2} \times \dfrac{3}{2}\\[5pt] &= \left( \dfrac{3}{2} \right)^2 \end{align*}

Now, we find the reciprocal of the base $\left(\dfrac{3}{2}\right),$ and then raise that reciprocal to the same power but with the opposite sign $(-2),$ as follows: \[ \left( \dfrac{3}{2} \right)^2 = \left(\dfrac23\right)^{-2} \]

First, we write $\dfrac{8}{125}$ as a base raised to a power:

\begin{align*} \dfrac {8} {125} &= \dfrac{2 \times 2 \times 2}{5 \times 5 \times 5}\\[5pt] &= \dfrac{2}{5} \times \dfrac{2}{5} \times \dfrac{2}{5}\\[5pt] &= \left( \dfrac{2}{5} \right)^3 \end{align*}

Now, we find the reciprocal of the base $\left(\dfrac{2}{5}\right),$ and then raise that reciprocal to the same power but with the opposite sign $(-3),$ as follows: \[ \left( \dfrac{2}{5} \right)^3 = \left(\dfrac 52\right)^{-3} \]