301 Moved Permanently - Free Printable

Let's solve each problem step-by-step using the rules of exponents and rational numbers.

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🔷 Key Rules Used:

1. $ a^{-n} = \frac{1}{a^n} $
2. $ \left(\frac{a}{b}\right)^{-n} = \left(\frac{b}{a}\right)^n $
3. $ \left(\frac{a}{b}\right)^{-1} = \frac{b}{a} $
4. $ (-a)^{-n} = \frac{1}{(-a)^n} $

We will express all answers in the form $ \frac{p}{q} $, where $ p $ and $ q $ are integers and $ q \ne 0 $.

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✔ Part 1: Express as rational number

#### (1) $ 2^{-3} $
$$
= \frac{1}{2^3} = \frac{1}{8}
\Rightarrow \boxed{\frac{1}{8}}
$$

#### (2) $ 3^{-3} $
$$
= \frac{1}{3^3} = \frac{1}{27}
\Rightarrow \boxed{\frac{1}{27}}
$$

#### (3) $ 4^{-3} $
$$
= \frac{1}{4^3} = \frac{1}{64}
\Rightarrow \boxed{\frac{1}{64}}
$$

#### (4) $ (-4)^{-2} $
$$
= \frac{1}{(-4)^2} = \frac{1}{16}
\Rightarrow \boxed{\frac{1}{16}}
$$

#### (5) $ (-2)^{-2} $
$$
= \frac{1}{(-2)^2} = \frac{1}{4}
\Rightarrow \boxed{\frac{1}{4}}
$$

#### (6) $ (-3)^{-2} $
$$
= \frac{1}{(-3)^2} = \frac{1}{9}
\Rightarrow \boxed{\frac{1}{9}}
$$

#### (7) $ \frac{1}{3^{-2}} $
Note: This is $ 1 \div 3^{-2} = 3^2 $
$$
= 3^2 = 9 = \frac{9}{1}
\Rightarrow \boxed{\frac{9}{1}}
$$

#### (8) $ \frac{1}{2^{-2}} $
$$
= 2^2 = 4 = \frac{4}{1}
\Rightarrow \boxed{\frac{4}{1}}
$$

#### (9) $ \frac{1}{5^{-2}} $
$$
= 5^2 = 25 = \frac{25}{1}
\Rightarrow \boxed{\frac{25}{1}}
$$

#### (10) $ \left(\frac{1}{2}\right)^{-5} $
$$
= \left(\frac{2}{1}\right)^5 = 2^5 = 32 = \frac{32}{1}
\Rightarrow \boxed{\frac{32}{1}}
$$

#### (11) $ \left(\frac{2}{3}\right)^{-2} $
$$
= \left(\frac{3}{2}\right)^2 = \frac{9}{4}
\Rightarrow \boxed{\frac{9}{4}}
$$

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✔ Part 2: Express as rational number

#### (12) $ 6^{-1} $
$$
= \frac{1}{6}
\Rightarrow \boxed{\frac{1}{6}}
$$

#### (13) $ 2^{-1} $
$$
= \frac{1}{2}
\Rightarrow \boxed{\frac{1}{2}}
$$

#### (14) $ 3^{-1} $
$$
= \frac{1}{3}
\Rightarrow \boxed{\frac{1}{3}}
$$

#### (15) $ 4^{-1} $
$$
= \frac{1}{4}
\Rightarrow \boxed{\frac{1}{4}}
$$

#### (16) $ (-7)^{-1} $
$$
= \frac{1}{-7} = -\frac{1}{7}
\Rightarrow \boxed{-\frac{1}{7}}
$$

#### (17) $ (-3)^{-1} $
$$
= \frac{1}{-3} = -\frac{1}{3}
\Rightarrow \boxed{-\frac{1}{3}}
$$

#### (18) $ (-5)^{-1} $
$$
= \frac{1}{-5} = -\frac{1}{5}
\Rightarrow \boxed{-\frac{1}{5}}
$$

#### (19) $ (-6)^{-1} $
$$
= \frac{1}{-6} = -\frac{1}{6}
\Rightarrow \boxed{-\frac{1}{6}}
$$

#### (20) $ \left(\frac{1}{4}\right)^{-1} $
$$
= \frac{4}{1} = 4
\Rightarrow \boxed{\frac{4}{1}}
$$

#### (21) $ \left(\frac{1}{2}\right)^{-1} $
$$
= \frac{2}{1} = 2
\Rightarrow \boxed{\frac{2}{1}}
$$

#### (22) $ \left(\frac{1}{3}\right)^{-1} $
$$
= \frac{3}{1} = 3
\Rightarrow \boxed{\frac{3}{1}}
$$

#### (23) $ \left(\frac{1}{5}\right)^{-1} $
$$
= \frac{5}{1} = 5
\Rightarrow \boxed{\frac{5}{1}}
$$

#### (24) $ (-4)^{-1} \times \left(\frac{-3}{2}\right)^{-1} $

Break it down:

- $ (-4)^{-1} = \frac{1}{-4} = -\frac{1}{4} $
- $ \left(\frac{-3}{2}\right)^{-1} = \frac{2}{-3} = -\frac{2}{3} $

Now multiply:
$$
-\frac{1}{4} \times -\frac{2}{3} = \frac{2}{12} = \frac{1}{6}
\Rightarrow \boxed{\frac{1}{6}}
$$

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✔ Final Answers Summary:

| Q.No | Answer |
|------|--------------|
| (1) | $ \frac{1}{8} $ |
| (2) | $ \frac{1}{27} $ |
| (3) | $ \frac{1}{64} $ |
| (4) | $ \frac{1}{16} $ |
| (5) | $ \frac{1}{4} $ |
| (6) | $ \frac{1}{9} $ |
| (7) | $ \frac{9}{1} $ |
| (8) | $ \frac{4}{1} $ |
| (9) | $ \frac{25}{1} $ |
| (10) | $ \frac{32}{1} $ |
| (11) | $ \frac{9}{4} $ |
| (12) | $ \frac{1}{6} $ |
| (13) | $ \frac{1}{2} $ |
| (14) | $ \frac{1}{3} $ |
| (15) | $ \frac{1}{4} $ |
| (16) | $ -\frac{1}{7} $ |
| (17) | $ -\frac{1}{3} $ |
| (18) | $ -\frac{1}{5} $ |
| (19) | $ -\frac{1}{6} $ |
| (20) | $ \frac{4}{1} $ |
| (21) | $ \frac{2}{1} $ |
| (22) | $ \frac{3}{1} $ |
| (23) | $ \frac{5}{1} $ |
| (24) | $ \frac{1}{6} $ |

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✔ All answers are expressed as rational numbers $ \frac{p}{q} $ with integers $ p, q $ and $ q \ne 0 $. Let me know if you'd like this formatted as a printable answer key!