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BECE Mathematics 2024 Paper

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Question : 20

Total: 40

Simplify:

\;\frac{27^{m+1}}{3^{m+2}}

$3^{2m}$
$3^{2m-1}$
$3^{2m+2}$
$3^{2m+1}$

Solution:

Express both the numerator and the denominator to be in the same base.

27=3 \times 3 \times 3=3^3

Apply

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

\;\;\frac{27^{m+1}}{3^{m+2}}=\;\frac{(3^3)^{m+1}}{3^{m+2}}

\;\;\frac{27^{m+1}}{3^{m+2}}=\;\frac{3^{3 \times (m+1)}}{3^{m+2}}

\;\;\frac{27^{m+1}}{3^{m+2}}=\;\frac{3^{3m+3}}{3^{m+2}}

Law of division of indices

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

\;\;\frac{27^{m+1}}{3^{m+2}}=3^{3m+3-(m+2)}

\;\;\frac{27^{m+1}}{3^{m+2}}=3^{3m+3-m-2}

\;\;\frac{27^{m+1}}{3^{m+2}}=3^{3m-m+3-2}

\;\;\frac{27^{m+1}}{3^{m+2}}=3^{2m+1}

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