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BECE Mathematics 2013 Paper

Question : 7

Total: 40

Simplify

\frac{2^2 \times 3^2}{4^2 \times 3^3}

Solution:

Method I

Using the law of indices

Change the

4^2

to base 2 so that the base 2 s can simplify each other.

4=2 \times 2

4^2=(2 \times 2)^2

But

(a^m \times b^n)^c=a^{m \times c} \times b^{n \times c}

2=2^1

(2 \times 2)^2=(2^1 \times 2^1)^2

(2 \times 2)^2=2^{1 \times 2} \times 2^{1 \times 2}

(2 \times 2)^2=2^2 \times 2^2

Law of multiplication of indices

a^m \times a^n=a^{m+m}

(2 \times 2)^2=2^{2+2}

(2 \times 2)^2=2^4

Alternate Solution

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

4=2 \times 2=2^2

4^2=(2^2)^2=2^{2 \times 2}=2^4

\frac{2^2 \times 3^2}{4^2 \times 3^3} = \frac{2^2 \times 3^2}{2^4 \times 3^3}

Law of division of indices

\frac{a^m}{a^n}=a^{m-n}

\frac{2^2 \times 3^2}{2^4 \times 3^3}=2^{2-4} \times 3^{2-3}=2^{-2} \times 3^{-1}

But

a^{-b}=\frac{1}{a^b}

2^{-2} \times 3^{-1}=\frac{1}{2^2} \times \frac{1}{3^1}

2^2=2 \times 2=4

3^1=3

2^{-2} \times 3^{-1}=\frac{1}{4} \times \frac{1}{3}=\frac{1 \times 1}{4 \times 3}=\frac{1}{12}

Method II

Change the indices to multiplication, cancel out and simplify

2^2=2 \times 2

4^2=4 \times 4

3^2=3 \times 3

3^3=3 \times 3 \times 3

\frac{2^2 \times 3^2}{4^2 \times 3^3}=\frac{2 \times 2 \times 3 \times 3}{4 \times 4 \times 3 \times 3 \times 3}=\frac{1 \times 1 \times 1 \times 1}{2 \times 2 \times 1 \times 1 \times 3}=\frac{1}{4 \times 3}=\frac{1}{12}

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