(a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\) (b) Without using M...

Assessment: WAEC SSCE - General Mathematics - 1993 (Essay) Subject: General Mathematics

Question 1 Report

(a) Simplify, without using Mathematical tables: \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\)

(b) Without using Mathematical tables, calculate \(\sqrt{\frac{P}{Q}}\) where \(P = 3.6 \times 10^{-3}\) and \(Q = 2.25 \times 10^{6}\), leaving your answer in standard form.

Answer Details

(a) Use \(a\log b = \log b^a\) and combine:

\[\log_{10}\frac{30}{16} - 2\log_{10}\frac{5}{9} + \log_{10}\frac{400}{243} = \log_{10}\frac{30}{16} - \log_{10}\frac{25}{81} + \log_{10}\frac{400}{243}\]

\[= \log_{10}\!\left(\frac{30}{16} \times \frac{81}{25} \times \frac{400}{243}\right)\]

Simplify the product step by step: \(\dfrac{30}{16} = \dfrac{15}{8}\), and \(\dfrac{15}{8}\times\dfrac{81}{25} = \dfrac{243}{40}\), then \(\dfrac{243}{40}\times\dfrac{400}{243} = \dfrac{400}{40} = 10\).

\[= \log_{10} 10 = 1\]

(b) With \(P = 3.6\times10^{-3}\) and \(Q = 2.25\times10^{6}\):

\[\frac{P}{Q} = \frac{3.6\times10^{-3}}{2.25\times10^{6}} = 1.6\times10^{-9} = 16\times10^{-10}\]

\[\sqrt{\frac{P}{Q}} = \sqrt{16\times10^{-10}} = 4\times10^{-5}\]

In standard form, \(\sqrt{\dfrac{P}{Q}} = 4.0\times10^{-5}\).

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