Number Forms (Indices And Logarithm) - SS1 Mathematics Past Questions and Answers - page 1

Solution (a) \(64^{\frac{5}{6}} = \ \sqrt[6]{64^{5}} = \ \sqrt[6]{{(2^{6})}^{5}} = \ \sqrt[6]{2^{6 \times 5}} = \ 2^{\frac{6 \times 5}{6}} = \ 2^{5} = 32\)

(b) \(40^{5} \div 40^{3} = \ 40^{5 - 3} = \ 40^{2} = 1600\)

(c) \({(5^{\frac{1}{- 2}})}^{4} = \ 5^{\frac{1}{- 2} \times \frac{4}{1}} = \ 5^{\frac{4}{- 2}} = \ 5^{- 2} = \ \frac{1}{5^{2}} = \ \frac{1}{25}\)

Solution (a) \({40a}^{4}b^{6}c^{3}\ \div \ - {8a}^{4}b^{3}c^{2}\ = \left( \frac{40}{- 8} \right)a^{4 - 4}b^{6 - 3}c^{3 - 2} = - 5a^{0}b^{3}c^{1} = - 5(1)b^{3}c = \ - 5b^{3}c\)

(b) \(- 2x^{3}\ \times \ {3x}^{- 2} = ( - 2 \times 3)x^{3 + - 2} = ( - 6)x^{3 - 2} = \ - 6x\)

\(x^{3}{\times x}^{7} = \ x^{3 + 7} = \ x^{10}\)

\({5mn}^{2} \times \left( {- 10m}^{3}n^{3}p \right) = (5 \times - 10)m^{1 + 3}n^{2 + 3}p = \ - 50m^{4}n^{5}p\)

(a) \(36380 = 3.6380\ \times 10000 = 3.6380\ \times \ 10^{4}\) (b) \(0.0000003427 = 3.427 \times 10^{- 7}\)

\(4^{x} = \frac{1}{64}\ \rightarrow \ 4^{x} = \ \frac{1}{2^{6}}\ \rightarrow \ 4^{x} = \ 2^{- 6}\ \rightarrow \ 2^{2x} = \ 2^{- 6},\ comparing\ powers,\ thus,\ 2x = \ - 6\ \rightarrow x = \frac{- 6}{2} = \ - 3\)

\(\log_{10}15 = \ \log_{10}(3 \times 5) = \ \log_{10}3 + \ \log_{10}5 = \ 0.4771 + 0.6989 = 1.176\)