Study S3 Mathematics Laws Of Indices - Geniebook

Laws Of Indices

Index Notation

Let us see how an index is represented

Consider a value, $$2^3$$

The way this is pronounced is “2 to the power 3”. The 2 is called the base.

The 3 is called the power or index.

• In singular, the terminology is index.
• In plural, the terminology is indices.

Question 1:

Evaluate $$2^3$$

Solution:

To evaluate this, write 2 three times and then multiply them.

\begin{align*} &= 2 \times 2 \times 2 \\ &= 8 \end{align*}

Question 2:

Write $$3 \times 3 \times 3 \times 3 \times 3$$ in index notation.

Solution:

The base is 3 and there are five 3’s.

Hence it can be written as,

$$3 \times 3 \times 3 \times 3 \times 3 = 3^5$$

Multiply it to get the answer.

$$= 243$$

Therefore the answer is $$243$$.

Law 1 Of Indices

If the base is the same, but has different powers in multiplication we can add up the powers or index.

$$a^m + a^n = a^{m+n}$$

Simplify each of the following leaving your answer in index notation where appropriate.

Question 3:

(a) $$6^2 \times 6^3$$

Solution:

By the law 1 of indices, the base is kept the same and powers are added.
\begin{align*} &=6^{2+3}\\ &= 6^5 \end{align*}

Therefore the answer is $$6^5$$

(b) $$(-2)^2 \times (-2)^5$$

Solution:

By the law 1 of indices,the base is kept the same and powers are added.

\begin{align*} &=(-2)^{2+5}\\ &= (-2)^7 \end{align*}

Therefore the answer is $$(-2)^7$$

Question 4:

Simplify each of the following.

A) $$x^2y \times x^3y^2$$

Solution:

The equation is in multiplication, hence it can be also written as,

\begin{align*} &=x^2 \times y \times x^3 \times y^2 \\ &=x^2 \times x^3 \times y \times y^2 \end{align*}

By applying law 1 of indices,

\begin{align*} &=x^{2+3} \times y^{1+2} \\ &= x^5y^3 \end{align*}

Therefore the answer is $$x^5y^3$$

B) $$-3pq \times 5p^2q^5$$

Solution:

The equation is in multiplication, hence it can be also written as,

\begin{align*} &=-3 \times p \times q \times 5 \times p^2 \times q^5\\ &=-3 \times 5 \times p \times p^2 \times q \times q^5 \end{align*}

By applying law 1 of indices

\begin{align*} &=-15 \times p^{1+2} \times q^{1+5}\\ &= -15p^3q^6 \end{align*}

Therefore the answer is $$-15p^3q^6$$

Law 2 Of Indices

If the base is the same, but has different powers in division, we can subtract the powers or index.

$$a^m \div a^n = a^{m-n}$$

Simplify each of the following, leaving your answer in index notation where appropriate.

Question 5:

A) $$10^5 \div 10^2$$

Solution:

The base is same, we are using law 2 and subtracting the powers,

\begin{align*} &=10^{5-2}\\ &=10^3 \end{align*}

Therefore the answer is $$10^3$$.

B) $$-(3)^7+(-3)^3$$

Solution:

The base is same, we are using law 2 and subtracting the powers,

\begin{align*} &=(-3)^{7-3}\\ &=(-3)^4 \end{align*}

Therefore the answer is $$(-3)^4$$

C) $$x^8 \div x^2$$

Solution:

The base is same, we are using law 2 and subtracting the powers

\begin{align*} &=x^{8-2}\\ &= x^6 \end{align*}

Therefore the answer is $$x^6$$

Question 6:

Simplify each of the following.

A) $$20\,p^5\,q^3 \,\div\, 5\,p^2\,q$$

Solution:

Dividing 20 by 5 and following the law 2 of indices for p and q

\begin{align*} &=4\,p^{5-2}\,q^{3-1}\\ &= 4\,p^3\,q^2 \end{align*}

Therefore the answer is $$4\,p^3\,q^2$$

Law 3 of Indices

If the base has a power which is multiplied by another power, the base should be the same and the power should be multiplied with each other.

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

Simplify each of the following leaving your answer in index notation where appropriate.

Question 7:

A) $$(3^4)^2$$

Solution:

By law 3,the base is kept the same and the powers are multiplied.

\begin{align*} &=3^{4\times2}\\ &=3^8 \end{align*}

Therefore the answer is $$3^8$$

B) $$(6^a)^2$$

Solution:

By law 3,the base is kept the same and the powers are multiplied.

\begin{align*} &=6^{a \times 2}\\ &= 6^{2a} \end{align*}

Therefore the answer is $$6^{2a}$$

C) $$(2^a)^4 \times (2^3)^a$$

Solution:

By law 3 of indices, the base is kept the same and the powers are multiplied

\begin{align*} &=2^{a\times4} \times 2^{3\times a}\\ &= 2^{4a} \times 2^{3a} \end{align*}

By law 1 of indices, the base is kept the same and powers are added

\begin{align*} &=2^{4a+3a}\\ &=2^{7a} \end{align*}

Therefore the answer is $$2^{7a}$$

Question 8:

Simplify each of the following.

A) $$(m^5)^2$$

Solution:

Using law 3, the base is kept the same and the powers are multiplied to get the answer.

\begin{align*} &=m^{5 \times 2}\\ &=m^{10} \end{align*}

Therefore the answer is $$m^{10}$$

B) $$(5^n)^3 \times (5^2)^n$$

Solution:

The powers for each value is multiplied,

$$= 5^{3n} \times 5^{2n}$$

Since it is in multiplication, we are using the law 1 of indices, the base is kept the same and the powers are added

\begin{align*} &=5^{3n+2n}\\ &=5^{7n} \end{align*}

Therefore the answer is $$5^{7n}$$

C) $$(7^h)^2 \times (7^5)^k$$

Solution:

The powers are multiplied with each other,

$$=7^{2h} \times 7^{5k}$$

Since it is in multiplication, we are using the law 1 of indices, the base is kept the same and the powers are added.

$$=7^{2h+5k}$$

Therefore the answer is $$=7^{2h+5k}$$

Law 4 of Indices

(i) Forward:

If the power is the same with the different bases, the bases can be multiplied and the power stays the same.

$$a^n \times b^n = (ab)^n$$

Simplify each of the following leaving your answer in index notation where appropriate.

Question 9:

A) $$4^3 \times 7^3$$

Solution:

Since the power is the same, the bases can be multiplied and the power remains the same.

\begin{align*} &=(4\times7)^{3}\\ &= 28^3 \end{align*}

Therefore the answer is $$28^3$$

B) $$(-8)^5 \times 3^5$$

Solution:

Since the power is the same, the bases can be multiplied and the power remains the same

\begin{align*} &=(-8 \times 3)^5\\ &= (-24)^5 \end{align*}

Therefore the answer is $$(-24)^5$$

C) $$p^2 \times q^2$$

Solution:

The power is the same here so we are multiplying the base and the power will remain the same

\begin{align*} &=(p \times q)^2\\ &= (pq)^2\\ &=p^2q^2 \end{align*}

Therefore the answer is $$p^2q^2$$

(ii) Reverse

It is the opposite of the previous law. When the base is a multiplied value of a number and a variable, the power or the index is common to both. So it can be written as the follows,

$$(ab)^n = a^n \times b^n$$

Simplify each of the following leaving your answer in index notation where appropriate.

D) $$(2a)^3$$

Solution:

Since the power is same for the value 2a, the question can be written as,

\begin{align*} &=2^3 \times a^3\\ &= (8a)^3 \end{align*}

$$8a^3$$

E) $$(-3b)^2$$

Solution:

Since the power is same for the value $$(-3b)$$, the question can be written as,

\begin{align*} &=(-3)^2 \times (b)^2\\ &= (9b)^2 \end{align*}

Therefore the answer is $$(9b)^2$$

F) $$(4p^2q)^3$$

Solution:

The common power 3 is multiplied with the values in the brackets,

$$=4^3 \times (p^2)^3 \times q^3$$

By law 4, the power of p , is multiplied to get the answers.

$$=64p^6q^3$$

Therefore the answer is $$64p^6q^3$$

Question 10:

Simplify each of the following.

A) $$a^5 \times 7^5$$

Solution:

The base 7 is written 5 times and multiplied to get the answer.

\begin{align*} &=16807\times a^5\\ &=16807a^5 \end{align*}

Therefore the answer is $$16807a^5$$

B) $$(3p^4)^2$$

Solution:

The common power 2 is multiplied with each variable in the bracket.

$$=3^2 \times (p^4)^2$$

By law 3, the powers of p are multiplied,

\begin{align*} &=9 \times p^8\\ &= 9p^8 \end{align*}

Therefore the answer is $$9p^8$$

C) $$(xy^2)^3 \times (-5x^4y)^2$$

Solution:

The common powers are multiplied with the values in the bracket.

$$=x^3 \times (y^2)^3 \times (-5)^2 \times (x^4)^2 \times y^2$$

By law 3, the powers are multiplied,

$$=25 \times x^3 \times x^8 \times y^6 \times y^2$$

By law 1, the powers of the same bases are added.

\begin{align*} &=25 \times x^{3+8} \times y^{6+2}\\ &=25x^{11}y^8 \end{align*}

Therefore the answer is $$25x^{11}y^8$$

Conclusion

In this article, we learnt about the Laws of indices and how to use those laws to solve various types of problems. The details covered are aligned to the syllabus of Secondary 3 Mathematics grade in Singapore.

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