(a)(i) Given that \(\log_{10} 5 = 0.699\) and \(\log_{10} 3 = 0.477\), find \(\log_{10} 45\), without using Mathematical tables. (ii) Hence, solve \(x^{0.82...

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

Question 1 Report

(a)(i) Given that \(\log_{10} 5 = 0.699\) and \(\log_{10} 3 = 0.477\), find \(\log_{10} 45\), without using Mathematical tables.

(ii) Hence, solve \(x^{0.8265} = 45\).

(b) Use Mathematical tables to evaluate \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\)

Answer Details

(a)(i) Write \(45 = 9 \times 5 = 3^2 \times 5\):

\[\log_{10}45 = 2\log_{10}3 + \log_{10}5 = 2(0.477) + 0.699 = 0.954 + 0.699 = 1.653\]

(ii) Solve \(x^{0.8265} = 45\). Take logarithms of both sides:

\[0.8265\,\log_{10}x = \log_{10}45 = 1.653\]

\[\log_{10}x = \frac{1.653}{0.8265} = 2.000 \;\Rightarrow\; x = 10^{2} = 100\]

(b) Evaluate \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}}\) using logarithms.

NumberLog
2.0670.3155
0.0348\(\bar{2}.5416\)
0.538\(\bar{1}.7308\)

Denominator log \(= \bar{2}.5416 + \bar{1}.7308 = \bar{2}.2724\). Then

\[\log(\text{fraction}) = 0.3155 - \bar{2}.2724 = 2.0431\]

\[\log(\text{answer}) = \tfrac12(2.0431) = 1.0216 \;\Rightarrow\; \text{answer} = 10.51\]

Hence \(\sqrt{\dfrac{2.067}{0.0348 \times 0.538}} \approx 10.51\).

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