Using the Index online exercise for - Free Printable

Let's solve each part of the Practice Sheet 1 on Indices (Exponents) step by step. We'll use the laws of indices:

🔷 Laws of Indices:

1. $ a^m \times a^n = a^{m+n} $
2. $ a^m \div a^n = a^{m-n} $
3. $ (a^m)^n = a^{m \times n} $
4. $ a^{-n} = \frac{1}{a^n} $
5. $ a^{0} = 1 $
6. $ a^{\frac{1}{n}} = \sqrt[n]{a} $
7. $ a^{\frac{m}{n}} = (\sqrt[n]{a})^m $

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## ✔ Question 1: Simplify and leave in index form

(a) $ 6^{-4} \times 6^7 $

Use law: $ a^m \times a^n = a^{m+n} $

$$
6^{-4 + 7} = 6^3
$$

✔ Answer: $ 6^3 $

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(b) $ 10^8 \times 10^{-5} $

$$
10^{8 + (-5)} = 10^3
$$

✔ Answer: $ 10^3 $

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(c) $ x^7 \times x^3 $

$$
x^{7+3} = x^{10}
$$

✔ Answer: $ x^{10} $

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(d) $ (x^{-2})^3 $

Use law: $ (a^m)^n = a^{m \times n} $

$$
x^{-2 \times 3} = x^{-6}
$$

✔ Answer: $ x^{-6} $

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(e) $ y^{-12} \times y^5 $

$$
y^{-12 + 5} = y^{-7}
$$

✔ Answer: $ y^{-7} $

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(f) $ y^8 \div y^3 $

$$
y^{8 - 3} = y^5
$$

✔ Answer: $ y^5 $

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(g) $ 7^2 \div 7^{-4} $

$$
7^{2 - (-4)} = 7^{2 + 4} = 7^6
$$

✔ Answer: $ 7^6 $

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(h) $ (m^4)^{-2} \times (m^3)^5 $

First simplify each part:

- $ (m^4)^{-2} = m^{4 \times (-2)} = m^{-8} $
- $ (m^3)^5 = m^{3 \times 5} = m^{15} $

Now multiply:

$$
m^{-8} \times m^{15} = m^{-8 + 15} = m^7
$$

✔ Answer: $ m^7 $

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(i) $ y^6 \times y^{14} \div y^5 $

Do multiplication first, then division:

$$
y^{6 + 14} = y^{20}, \quad y^{20} \div y^5 = y^{20 - 5} = y^{15}
$$

✔ Answer: $ y^{15} $

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(j) $ (8^3)^4 \div (8^2)^3 $

Simplify each part:

- $ (8^3)^4 = 8^{3 \times 4} = 8^{12} $
- $ (8^2)^3 = 8^{2 \times 3} = 8^6 $

Now divide:

$$
8^{12} \div 8^6 = 8^{12 - 6} = 8^6
$$

✔ Answer: $ 8^6 $

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## ✔ Question 2: Simplify the following

(a) $ 9^{\frac{1}{2}} $

$ 9 = 3^2 $, so:

$$
(3^2)^{\frac{1}{2}} = 3^{2 \times \frac{1}{2}} = 3^1 = 3
$$

✔ Answer: $ 3 $

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(b) $ 27^{\frac{1}{3}} $

$ 27 = 3^3 $, so:

$$
(3^3)^{\frac{1}{3}} = 3^{3 \times \frac{1}{3}} = 3^1 = 3
$$

✔ Answer: $ 3 $

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(c) $ 16^{\frac{1}{2}} $

$ 16 = 4^2 = 2^4 $, so:

$$
(2^4)^{\frac{1}{2}} = 2^{4 \times \frac{1}{2}} = 2^2 = 4
$$

✔ Answer: $ 4 $

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(d) $ 16^{-\frac{1}{2}} $

We know $ 16^{\frac{1}{2}} = 4 $, so:

$$
16^{-\frac{1}{2}} = \frac{1}{16^{\frac{1}{2}}} = \frac{1}{4}
$$

✔ Answer: $ \frac{1}{4} $

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(e) $ 27^{-\frac{2}{3}} $

First, $ 27^{\frac{1}{3}} = 3 $, so:

$$
27^{\frac{2}{3}} = (27^{\frac{1}{3}})^2 = 3^2 = 9
$$

Then:

$$
27^{-\frac{2}{3}} = \frac{1}{27^{\frac{2}{3}}} = \frac{1}{9}
$$

✔ Answer: $ \frac{1}{9} $

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## ✔ Rewrite the following in index form:

(a) $ \sqrt{8} $

$$
\sqrt{8} = 8^{\frac{1}{2}}
$$

✔ Answer: $ 8^{\frac{1}{2}} $

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(b) $ \sqrt[4]{m} $

$$
\sqrt[4]{m} = m^{\frac{1}{4}}
$$

✔ Answer: $ m^{\frac{1}{4}} $

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(c) $ (m^6)^{\frac{1}{2}} $

This is already in index form, but we can simplify:

$$
m^{6 \times \frac{1}{2}} = m^3
$$

But the question says rewrite in index form, so either $ (m^6)^{\frac{1}{2}} $ or $ m^3 $. Since it’s asking to rewrite, probably they want simplified index form.

✔ Answer: $ m^3 $

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(d) $ (10^{\frac{1}{2}})^3 $

$$
= 10^{\frac{1}{2} \times 3} = 10^{\frac{3}{2}}
$$

✔ Answer: $ 10^{\frac{3}{2}} $

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(e) $ (16^{\frac{1}{2}})^{-2} $

First, $ 16^{\frac{1}{2}} = 4 $, but let's keep it in index form:

$$
(16^{\frac{1}{2}})^{-2} = 16^{\frac{1}{2} \times (-2)} = 16^{-1}
$$

Alternatively, $ 16^{-1} = \frac{1}{16} $, but since it asks for index form, $ 16^{-1} $ is fine.

✔ Answer: $ 16^{-1} $

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## ✔ Final Answers Summary:

Question 1:

(a) $ 6^3 $
(b) $ 10^3 $
(c) $ x^{10} $
(d) $ x^{-6} $
(e) $ y^{-7} $
(f) $ y^5 $
(g) $ 7^6 $
(h) $ m^7 $
(i) $ y^{15} $
(j) $ 8^6 $

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Question 2: Simplify

(a) $ 3 $
(b) $ 3 $
(c) $ 4 $
(d) $ \frac{1}{4} $
(e) $ \frac{1}{9} $

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Rewrite in index form:

(a) $ 8^{\frac{1}{2}} $
(b) $ m^{\frac{1}{4}} $
(c) $ m^3 $
(d) $ 10^{\frac{3}{2}} $
(e) $ 16^{-1} $

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