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1993 WAEC Mathematics Theory (a)(i) Given that \(\log_{10} 5 = 0.699\) and \(\log_{10} 3 = 0.477\), find \(\log_{10} 45\),...

Mathematics
WAEC 1993

(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 \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\)

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Explanation

(a)(i) \(\log_{10} 45 = \log_{10} (3 \times 3 \times 5)\)

= \(\log_{10} (3^{2} \times 5)\)

= \(\log_{10} 3^{2} + \log_{10} 5\)

= \(2 \log_{10} 3 + \log_{10} 5\)

= \(2(0.477) + 0.699\)

= \(0.954 + 0.699 = 1.653\)

(ii) \(x^{0.8265} = 45\)

Taking the log of both sides,

\(\log_{10} x^{0.8265} = \log_{10} 45\)

\(0.8265 \log_{10} x = \log_{10} 45\)

\(\log_{10} x = \frac{1.653}{0.8265}\)

\(\log_{10} x = 2\)

\(x = 10^{2} = 100\)

(b) \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\)

No Log
2.067 \(0.0348\)       = 0.3513 -
0.0348

\(\bar{2}.5416 +\)

0.538 \(\bar{1}.7308\)
  \(\bar{2}.2724\) = \(\bar{2}.2724\)
                           = \(2.0789 \div 2 = 1.0395\)
Antilog - 10.95  

\(\therefore \sqrt{\frac{2.067}{0.0348 \times 0.538}} = 10.95\)

 


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