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WAEC Further Mathematics 2023 Paper

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Question : 17

Total: 40

\text{Simplify: }\frac{\log \sqrt{27} - \log \sqrt{8}}{\log 3 - \log 2}

$\frac{3}{2}$
- $\frac{1}{4}$
- $\frac{3}{2}$
$\frac{1}{4}$

Solution:

Concept:

Use the power rule of logarithms:

\log a^n = n \log a

.

Explanation:

Write the radicals as powers.

\sqrt{27} = \sqrt{3^3} = 3^{3/2}

and

\sqrt{8} = \sqrt{2^3} = 2^{3/2}

.

So,

\log \sqrt{27} - \log \sqrt{8} = \log 3^{3/2} - \log 2^{3/2}

.

Apply the power rule:

= \frac{3}{2}\log 3 - \frac{3}{2}\log 2

.

Factor out

\frac{3}{2}

:

= \frac{3}{2}(\log 3 - \log 2)

.

Thus the expression becomes

\frac{\frac{3}{2}(\log 3 - \log 2)}{\log 3 - \log 2}

.

Cancel

(\log 3 - \log 2)

from numerator and denominator.

The result is

\frac{3}{2}

.

Answer:

A.

\frac{3}{2}

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