(a) \(\frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}}\)
\(\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8}) = \frac{3}{4}(\frac{27}{8} + \frac{13}{8})\)
= \(\frac{3}{4}(\frac{40}{8})\)
= \(\frac{15}{4}\)
\(2\frac{1}{8} - 1\frac{1}{2} = \frac{17}{8} - \frac{3}{2}\)
= \(\frac{17}{8} - \frac{12}{8}\)
= \(\frac{5}{8}\)
\(\therefore \frac{\frac{3}{4}(3\frac{3}{8} + 1\frac{5}{8})}{2\frac{1}{8} - 1\frac{1}{2}} = \frac{15}{4} \div \frac{5}{8}\)
= \(\frac{15}{4} \times \frac{8}{5}\)
= \(6\)
(b) \(\log_{10} 2 = 0.3010 ; \log_{10} 3 = 0.4771\)
\(\log_{10} 1.125 = \log_{10}(\frac{1125}{1000})\) (Dividing through with 125)
= \(\log_{10} (\frac{9}{8})\)
= \(\log_{10} 9 - \log_{10} 8\)
\(\log_{10} 9 = \log_{10} 3^{2} = 2\log_{10} 3 = 2 \times 0.4771 = 0.9542\)
\(\log_{10} 8 = \log_{10} 2^{3} = 3\log_{10} 2 = 3 \times 0.3010 = 0.9030\)
\(\therefore \log_{10} (\frac{9}{8}) = 0.9542 - 0.9030 = 0.0512\)
\(\approxeq 0.051\) ( 2 significant figures)