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(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two. I'll be using the latter method. \(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\) = \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\) \(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \) = \(4 + 0 + 1 = 5_{ten}\) \(\frac{255}{5} = 51_{ten}\) 2 51 2 25 r 1 2 12 r 1 2 6 r 0 2 3 r 0 2 1 r 1 0 r 1 = \(110011_{two}\) (b) (i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\) = \(11cm\) \(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\) (ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\) = \(33 cm^{2}\)

(a) \(\log_{10} (3x - 1) - \log_{10} 2 = 3\) \(\log_{10} (\frac{3x - 1}{2}) = 3\) \(\implies \frac{3x - 1}{2} = 10^{3} = 1000\) \(\implies 3x - 1 = 2000\) \(3x = 2000 + 1 = 2001\) \(x = \frac{2001}{3} = 667\) (b) No Log 0.8972 \(\bar{1}.9528\) + 3.536 \(0.5485\) = 0.5013 - 0.00249 \(\bar{3}.3962\) Antilog - 35.70 \(3.1051 \div 2 = 1.5526\) \(\therefore \sqrt{\frac{0.8972 \times 3.536}{0.00249}} = 35.70\)

TH = tan 45o, TH = QH \(\frac{TH}{5 + QH}\) = tan 30o TH = (b + QH) tan 30o QH = 56 (5 + QH) \(\frac{1}{\sqrt{3}}\) QH(1 - \(\frac{1}{\sqrt{3}}\)) = \(\frac{5}{\sqrt{3}}\) QH = \(\frac{5\sqrt{3}}{\sqrt{3}} - \frac{1}{\sqrt{3}}\) = \(\frac{5}{\sqrt{3} - 1}\)

\(\frac{4^{-\frac{1}{2}} \times 16^{\frac{3}{4}}}{4^{\frac{1}{2}}}\) = \(\frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{(2^4)^{\frac{3}{4}}}{4^{\frac{1}{2}} \times 4^{\frac{1}{2}}}\) = \(\frac{2^3}{4}\) = 2

\(\frac{\log \sqrt{27}}{\log 81}\) = \(\frac{\log \sqrt{3^3}}{\log 3^4}\) = \(\frac{\log 3^{\frac{3}{2}}}{\log 3^4}\) = \(\frac{\frac{3}{2} \log 3}{4 \log 3}\) = \(\frac{\frac{3}{2}}{4}\) = \(\frac{3}{8}\)

9(x - \(\frac{1}{2}\)) 3x2 = 3\(^{2(x - \frac{1}{2})}\) = 3\(^x{^2}\) ∴ 2(x - \(\frac{1}{2}\)) = x2 2x - 1 = x\(^2\) hence x\(^2\) - 2x + 1 = 0 (x - 1)(x - 1) = 0 x = 1

10\(^y\) x 5\(^{(2y - 2)}\) x 4\(^{(y - 1)}\) = 1 but 10\(^y\) = (5 x 2)\(^y\) = 5\(^y\) x 2\(^y\) = (Law of indices) 5\(^y\) x 2\(^y\) x 5\(^{(2y - 2)}\) x 4\(^{(y - 1)}\) = 1 but 4\(^{(y - 1)}\) = 2\(^{2(y - 1)}\) = 2\(^{2y - 2}\) (Law of indices) 5\(^y\) x 5\(^{(2y -2)}\) x 2\(^{(2y- 2)}\) x 2\(^y\) = 1 5\(^{(3y -2)}\) x 2\(^y\) x 2\(^{(2y -2)}\) = 1 = 5\(^{(3y -2)}\) x 2\(^{(3y -2)}\) = 1 But a\(^o\) = 1 10\(^{(3y -2)}\) = 10\(^o\) 3y - 2 = 0 ∴ y = \(\frac{2}{3}\)

2log3y + log3x2 = 4 log3y2 + log3x2 = 4 ∴ log3 (x2y2) = log381(correct all to base 4) x2y2 = 81 ∴ xy = \(\pm\)9 ∴ y = \(\pm\)\(\frac{9}{x}\)

Let each of the unknown side be x. \(10^{2} = x^{2} + x^{2} - 2(x)(x) \cos 120\) \(100 = 2x^{2} - 2x^{2} \cos 120\) \(100 = 2x^{2} + x^{2} = 3x^{2}\) x = \(\sqrt{\frac{100}{3}}\) = \(\frac{10}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}\) x = \(\frac{10\sqrt{3}}{3}\)cm

SQ2 + OS2 = OQ2 + 52 = 142 SQ2 = 142 - 52 196 - 25 = 171 ST2 + TQ2 = SQ2 ST2 + 22 = 171 ST2 = 171 - 4 = 167 ST = \(\sqrt{167}\) = 12.92 = 12.9cm

(a) \(\log_{10} (\frac{30}{16}) - 2 \log_{10} (\frac{5}{9}) + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16}) - \log_{10} (\frac{5}{9})^{2} + \log_{10} (\frac{400}{243})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \div \frac{25}{81})\) = \(\log_{10} (\frac{30}{16} \times \frac{400}{243} \times \frac{81}{25})\) = \(\log_{10} (10) = 1\) (b) \(\sqrt{\frac{P}{Q}} = \sqrt{\frac{3.6 \times 10^{-3}}{2.25 \times 10^{6}}\) = \(\sqrt{\frac{36 \times 10^{-4}}{2.25 \times 10^{6}}\) = \(\frac{6 \times 10^{-2}}{1.5 \times 10^{3}}\) = \(4 \times 10^{-2 - 3} = 4 \times 10^{-5}\)

(a)(i) \(\log_{10} 45 = \log_{10} (3 \times 3 \times 5)\) = \(\log_{10} (3^{2} \times 5)\) = \(\log_{10} 3^{2} + \log_{10} 5\) = \(2 \log_{10} 3 + \log_{10} 5\) = \(2(0.477) + 0.699\) = \(0.954 + 0.699 = 1.653\) (ii) \(x^{0.8265} = 45\) Taking the log of both sides, \(\log_{10} x^{0.8265} = \log_{10} 45\) \(0.8265 \log_{10} x = \log_{10} 45\) \(\log_{10} x = \frac{1.653}{0.8265}\) \(\log_{10} x = 2\) \(x = 10^{2} = 100\) (b) \(\sqrt{\frac{2.067}{0.0348 \times 0.538}}\) No Log 2.067 \(0.0348\) = 0.3513 - 0.0348 \(\bar{2}.5416 +\) 0.538 \(\bar{1}.7308\) \(\bar{2}.2724\) = \(\bar{2}.2724\) = \(2.0789 \div 2 = 1.0395\) Antilog - 10.95 \(\therefore \sqrt{\frac{2.067}{0.0348 \times 0.538}} = 10.95\)

\(\log_{10} 25 + \log_{10} 32 - \log_{10} 8\) = \(\log_{10} (\frac{25 \times 32}{8})\) = \(\log_{10} 100 \) = 2

Let \(\log_{b} a^{n} = x\) \(\therefore a^{n} = b^{x}\) \(a^{n} = (a^{\frac{1}{n}})^{x}\) \(a^{n} = a^{\frac{x}{n}} \implies n = \frac{x}{n}\) \(x = n^{2}\)

\(\frac{4^{2x}}{4^{3x}}\) = 2 42x - 3x = 2 4-x = 2 (22)-x = 21 Equating coefficients: -2x = 1 x = -\(\frac{1}{2}\)