Given that \(\log_{10} 2 = 0.3010\) and \(\log_{10} 3 = 0.4771\), calculate without using mathematical tables or calculator, the value of :
(a) \(\log_{10} 54\) ;
(b) \(\log_{10} 0.24\).
(a) \(\log_{10} 54\) ;
(b) \(\log_{10} 0.24\).
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Correct Answer: Option n
Explanation:
(a) \(\log_{10} 54 \)
\(54 = 2 \times 3^{3}\)
\(\log_{10} 54 = \log_{10} (2 \times 3^{3})\)
= \(\log_{10} 2 + \log_{10} 3^{3}\)
= \(\log_{10} 2 + 3\log_{10} 3\)
= \(0.3010 + (3 \times 0.4771)\)
= \(0.3010 + 1.4313\)
= \(1.7323\)
(b) \(\log_{10} 0.24\)
\(0.24 = \frac{24}{100} = \frac{2^{3} \times 3}{100}\)
\(\log_{10} 0.24 = \log_{10} 24 - \log_{10} 100\)
\(\log_{10} (2^{3} \times 3) - \log_{10} (10^{2})\)
\(\log_{10} 2^{3} + \log_{10} 3 - 2\log_{10} 10\)
= \(3\log_{10} 2 + \log_{10} 3 - 2\)
= \((3 \times 0.3010) + 0.4771 - 2\)
= \(0.9030 + 0.4771 - 2\)
= \(-0.6199\)
(a) \(\log_{10} 54 \)
\(54 = 2 \times 3^{3}\)
\(\log_{10} 54 = \log_{10} (2 \times 3^{3})\)
= \(\log_{10} 2 + \log_{10} 3^{3}\)
= \(\log_{10} 2 + 3\log_{10} 3\)
= \(0.3010 + (3 \times 0.4771)\)
= \(0.3010 + 1.4313\)
= \(1.7323\)
(b) \(\log_{10} 0.24\)
\(0.24 = \frac{24}{100} = \frac{2^{3} \times 3}{100}\)
\(\log_{10} 0.24 = \log_{10} 24 - \log_{10} 100\)
\(\log_{10} (2^{3} \times 3) - \log_{10} (10^{2})\)
\(\log_{10} 2^{3} + \log_{10} 3 - 2\log_{10} 10\)
= \(3\log_{10} 2 + \log_{10} 3 - 2\)
= \((3 \times 0.3010) + 0.4771 - 2\)
= \(0.9030 + 0.4771 - 2\)
= \(-0.6199\)