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\(413_7\) to base 5 convert first to base 10 \(417_7 = 4 \times 7^2 + 1 \times 7^1 + 3 \times 7^0\) = 4 \times 49 + 1 \times 7 + 3 \times 1 = 196 + 7 + 3 = \(206_{10}\) convert this result to base 5 5 206 5 41R1 5 8R1 5 1R3 0R1 \(∴ 413_7 = 1311_5\)

Converting each number to base 10 \(110_{two} = 1 \times 2^2 + 1 \times 2^1 + 0 \times 2^0\) = 1 \times 4 + 1 \times 2 + 0 \times 1 = 4 + 2 + 0 = \(6_{ten}\) \(31_{eight} = 3 \times 8^1 + 1 \times 8^0\) = 3 \times 8 + 1 \times 1 = 24 + 1 = \(25_{ten}\) \(42_{five} = 4 \times 5^1 + 2 \times 5^0\) = 4 \times 5 + 2 \times 1 = 20 + 2 = \(22_{ten}\) Hence, \(31_{eight} > 42_{five} > 110_{two}\) In ascending order, \(110_{two}, 42_{five}, 31_{eight}\)

16\(^{0.16}\) \times 16\(^{0.04}\) \times 2\(^{0.2}\) =2\(^{4(0.16)}\) \times 2\(^{4(0.04)}\) \times 2\(^{0.2}\) =2\(^{0.64}\) \times 2\(^{0.16}\) \times 2\(^{0.2}\) =2\(^{0.64\times0.16\times0.2}\) =2\(^1\) 2

\(2^{5x} \div 2^x = \sqrt[5]{2^{10}}\) applying the laws of indices \(2^{5x - x} = 2^{10(1/5)}\) \(2^{4x} = 2^{10(1/5)}\) \(2^{4x} = 2^2\) Equating the powers then 4x = 2 therefore, x = \(\frac{2}{4}\) = \(\frac{1}{2}\)

log (y + 8) + log (y - 8) = 2log 3 + 2log 5 ⇒ log (y + 8)(y - 8) = 2log 3 + 2log 5 (log ab = log a + log b) ⇒ log (y\(^2\) -8y + 8y - 64) = log 3\(^2\) + log 5\(^2\) ⇒ log (y\(^2\) - 64) = log 3\(^2\) + log 5\(^2\) ⇒ log (y\(^2\) - 64) = log 9 + log 25 ⇒ log (y\(^2\) - 64) = log (9 \times 25) ⇒ log(y\(^2\) - 64) = log 225 ⇒ y\(^2\) - 64 = 225 ⇒ y\(^2\) = 225 + 64 ⇒ y\(^2\) = 289 ⇒ y = \pm\sqrt289 ∴ y = \pm17

\(log_2 (6 - x) = 3 - log_2 x\) ⇒ \(log_2 (6 - x) = 3 log_2 2 - log_2 x\) (since \(log_2\) 2 = 1) ⇒ \(log_2 (6 - x) = log_2 2^3 - log_2 x\) \((a log\) c = \(log\) c\(^a)\) ⇒ \(log_2 (6 - x) = log_2 8 - log_2 x\) ⇒\(log_2 (6 - x) = log_2 \frac{8}{x}\) (\(log\) a - \(log\) b = \(log \frac{a}{b})\) ⇒ \(6 - x = \frac{8}{x}\) ⇒ \(x (6 - x) = 8\) ⇒ \(6x - x^2 = 8\) ⇒ \(x^2 - 6x + 8 = 0\) ⇒ \(x^2 - 4x - 2x + 8 = 0\) ⇒ \(x (x - 4) - 2(x - 4) = 0\) ⇒ \((x - 4)(x - 2) = 0\) ⇒ \(x - 4 = 0 or x - 2 = 0\) ∴ x = 4 or 2

\(log_3 x + log_3 (x - 8) = 2\) \(log_3 x(x - 8) = log_39\) since 2 = \(log_39\) \(log_3\) cancels out ⇒ x(x - 8) = 9 ⇒ \(x^2 - 8x = 9\) ⇒ \(x^2 - 8x - 9 = 0\) ⇒ \(x^2 - 9x + x - 9 = 0\) ⇒ x(x - 9) + 1(x - 9) = 0 ⇒ (x - 9)(x + 1) = 0 ⇒ x = 9 or x = -1 Since we can't have a log of negative numbers, ∴ x = 9.

\(\frac {3 - \sqrt 3}{2 + \sqrt 3} = a + b\sqrt 3\) Rationalize = \(\frac {3 - \sqrt 3}{2 + \sqrt 3} \times \frac {2 - \sqrt 3}{2 - \sqrt 3}\) = \(\frac {(3 - \sqrt 3)}{(2 + \sqrt 3)} \frac {(2 - \sqrt 3)}{(2 - \sqrt 3)}\) = \(\frac {6 - 3 \sqrt 3 - 2 \sqrt 3 + (\sqrt 3)^2}{4 - 2 \sqrt 3 + 2 \sqrt 3 - (\sqrt 3)^2}\) = \(\frac {6 - 5 \sqrt 3 + 3}{4 - 3}\) = \(\frac {9 - 5 \sqrt 3}{1} = 9 - 5 \sqrt 3\) = 9 + (-5) \(\sqrt 3\) \(\therefore a = 9, b = - 5\)

\(3\sqrt{12} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(3\sqrt{4\times3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) = \(6\sqrt{3} + 10\sqrt{3} - \frac{6}{\sqrt{3}}\) treating like a fraction, then = \(\frac{18 + 30 - 6}{\sqrt{3}}\) = \(\frac{42}{\sqrt{3}}\) rationalizing = \(14\sqrt{3}\)

Original population = 1,230 New population = 1,040 Decrease in population = 1,230 – 1,040 = 190 Percentage decrease in population = decrease in population x 100% original population = \(\frac {190}{1,230}\) x 100 = 15.45%

Let the total number of items in the man's shop = \(y\) Number of Brand A's items in the man's shop = \(\frac {1}{9} y\) Remaining items = 1 - \(\frac {1}{9} y = \frac {8}{9} y\) Number of Brand B's items in The man's shop = \(\frac{5}{8} of \frac{8}{9}y = \frac{5}{9}y\) Total of Brand A and Brand B's items = \(\frac{1}{9}y + \frac{5}{9}y = \frac{2}{3}y\) Number of Brand C's items in the man's shop = 1 - \(\frac{2}{3}y = \frac{1}{3}y\) \(\implies\frac{1}{3}y\) = 81 (Given) \(\implies y\) = 81 x 3 = 243 ∴ The total number of items in the man's shop = 243 ∴ Number of Brand B's items in the man's shop = \(\frac{5}{9}\) x 243 = 135 ∴ The number of more Brand B items than Brand C = 135 - 81 =54

16.54 x 10\(^{-5}\) - 6.76 x 10\(^{-8}\) + 0.23 x 10\(^{-6}\) ⇒ 1.654 x 10\(^{-4}\) - 6.76 x 10\(^{-8}\) + 2.3 x 10\(^{-7}\) ⇒ 1.654 x 10\(^{-4}\) - 0.000676 x 10\(^{-4}\) + 0.0023 x 10\(^{-4}\) ⇒ (1.654 - 0.000676 + 0.0023) x 10\(^{-4}\) ∴ 1.655624 x 10\(^{-4}\) ≃ 1.66 x 10\(^{-4}\)

% error = \(\frac{Error}{Actual Value}\) x 100% Student's Value = 18mL Actual Value = 18.4mL Error = 18.4 - 18 = 0.4 ∴ % error = \(\frac{0.4}{18.4}\) x 100% = 2.17%

\(\frac{5}{8} - \frac{3}{4} \div \frac{5}{12} x \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{3}{4} \div \frac{5}{12}) \times \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{3}{4} \times \frac{12}{5}) \times \frac{1}{4}\) ⇒ \(\frac{5}{8} - (\frac{9}{5} \times \frac{1}{4})\) ⇒ \(\frac{5}{8} - \frac{9}{20}\) ∴ \(\frac{7}{40}\)

The two trailing zeros in the number are not significant, but the other five are, making it a five-figure number.