Question Library

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 = \(14\sqrt{3}\)

hyp = 5cm, adj = \(\frac{8cm}{2}\) 4cm, opp = xcm Pythagoras theorem: hyp\(^2\) = opp\(^2\) + adj\(^2\) 5\(^2\) = x\(^2\) + 4\(^2\) x\(^2\) = 25 - 16 x = \sqrt9 x = 3cm

\((8^{-1})^{k+2}\) = \((8^{0})\) base 8 cancel out on both sides -1(k+2) = 0 -k -2 = 0 : k = -2

4\(^{3x}\) = 4\(^{2(x+1)}\) 3x = 2x + 2 3x - 2x = 2 x = 2

where Log\(_2\) 8\sqrt2 → Log\(_2\) \sqrt128 → Log\(_2\) 128\(^\frac{1}{2}\) = \(\frac{1}{2}\) * (Log\(_2\) 128) → \(\frac{1}{2}\) * (Log\(_2\) 2\(^7\)) = 7 * \(\frac{1}{2}\) * (Log\(_2\) 2) where (Log\(_2\) 2) = 1 → 7 * \(\frac{1}{2}\) * 1 = \(\frac{7}{2}\) or 3.5

Recall that: log\(_3\) 27 → log\(_3\)3\(^3\) 3log\(_3\)3 → 3 * 1 = 3 Then log\(_3\) 27 = 2x + 1 → 3 = 2x + 1 3 - 1 = 2x 2 = 2x 1 = x

Rationalization with conjugate \({3+\sqrt2}\) \(\frac{1}{3-\sqrt2}\) → \(\frac{1 * [3+\sqrt2]}{[3-\sqrt2][3+\sqrt2]}\) = \(\frac{3+\sqrt2}{9 -3\sqrt2 + 3\sqrt2 - \sqrt4}\) = \(\frac{3+\sqrt2}{9 - 2}\) → \(\frac{3+\sqrt2}{7}\) = \(\frac{3}{7}\) + \(\frac{\sqrt2}{7}\)

\sqrt24 + \sqrt96 - \sqrt600 = y\sqrt6 convert all to basic surds \sqrt24 = 2\sqrt6 \sqrt96 = 4\sqrt6 - \sqrt600 = -10\sqrt6 2\sqrt6 + 4\sqrt6 - 10\sqrt6 = -4\sqrt6 value of y = -4

(a) y =(\(\frac{pr}{m} - p^2r\))\(^{\frac{-3}{2}}\) y = (\(\frac{pr - mrp^2}{m})^\frac{-3}{2}\) y = [\(\frac{m}{pr - mrp^2}]^\frac{3}{2}\) \(y^\frac{2}{3} = \frac{m}{r(p - mp^2)}\) r = \(\frac{m}{y^\frac{2}{3}[p - mp^2]}\) b) Given y = -8, m = 1 p = 3 r = \(\frac{m}{y^\frac{2}{3}[p - mp^2]}\) r = \(\frac{1}{(-8)^\frac{2}{3}[ 3 - (1)3^2]}\) r = \(\frac{1}{(-2)^2[3 - 9]}\) r = \(\frac{1}{[4][-6]}\) = \(\frac{-1}{24}\)

PQ\(^2\) = (x2 - x1)\(^2\) + (y2 - y1)\(^2\) = 12\(^2\) + 5\(^2\) = 144 + 25 = 169 PQ = \sqrt169 = 13 But PQ = diameter = 2r, r = PQ/2 = 6.5 units

first work out the expression and then correct the answer to 4 s.f = 241.34..............(A) (3 x 10-\(^3\))\(^2\)............(B) = 3\(^2\)x\(^2\) = \(\frac{1}{10^3}\) x \(\frac{1}{10^3}\) (Note that x\(^2\) = \(\frac{1}{x^3}\)) = 24.34 x 3\(^2\) x \(\frac{1}{10^6}\) = \(\frac{2172.06}{10^6}\) = 0.00217206 = 0.002172(4 s.f)