(a)(i) \(\log_{10} 35 = \log_{10} (7 \times 5) = \log_{10} 7 + \log_{10} 5\)
= \(0.8451 + 0.6990 = 1.5441\)
(ii) \(\log_{10} 2.8 = \log_{10} (\frac{4 \times 7}{10} = \log_{10} 4 + \log_{10} 7 - \log_{10} 10\)
= \(\log_{10} 2^{2} + \log_{10} 7 - 1\)
= \(2\log_{10} 2 + \log_{10} 7 - 1 = 2(0.3010) + 0.8451 - 1\)
= \(0.4471\)
(b) \(N^{0.8942} = 2.8\)
\(0.8942 \log N = \log 2.8\)
\(\log N = \frac{\log 2.8}{0.8942} = \frac{0.4471}{0.8942}\)
\(\log N = 0.5\)