(a) Given that \(\log_{10} 2 = 0.3010, \log_{10} 7 = 0.8451\) and \(\log_{10} 5 = 0.6990\), evaluate without using logarithm tables: (i) \(\log_{10} 35\); (...
Question 1 Report
(a) Given that \(\log_{10} 2 = 0.3010, \log_{10} 7 = 0.8451\) and \(\log_{10} 5 = 0.6990\), evaluate without using logarithm tables:
(i) \(\log_{10} 35\); (ii) \(\log_{10} 2.8\).
(b) Given that \(N^{0.8942} = 2.8\), use your result in (a)(ii) to find the value of N.
Answer Details
(a)(i) \(35 = 5\times7\), so by the addition law of logarithms:
\[\log_{10}35 = \log_{10}5 + \log_{10}7 = 0.6990 + 0.8451 = 1.5441.\]
(a)(ii) \(2.8 = \dfrac{28}{10} = \dfrac{4\times7}{10} = \dfrac{2^2\times7}{10}\):
\[\log_{10}2.8 = 2\log_{10}2 + \log_{10}7 - \log_{10}10 = 2(0.3010) + 0.8451 - 1 = 0.4471.\]
(b) Take logarithms of \(N^{0.8942} = 2.8\):
\[0.8942\log_{10}N = \log_{10}2.8 = 0.4471 \Rightarrow \log_{10}N = \frac{0.4471}{0.8942} = 0.5.\]
\[N = 10^{0.5} = \sqrt{10} = 3.16.\]
Download The App On Google Playstore
Everything you need to excel in your exams
