(i)

\begin{align} (\sqrt{5} - 2)x & = \sqrt{5} + 2 \\ x & = {\sqrt{5} + 2 \over \sqrt{5} - 2} \\ & = {\sqrt{5} + 2 \over \sqrt{5} - 2} \times {\sqrt{5} + 2 \over \sqrt{5} + 2} \phantom{00000000} [\text{Rationalise denominator}] \\ & = {(\sqrt{5} + 2)^2 \over (\sqrt{5} - 2)(\sqrt{5} + 2)} \\ & = {(\sqrt{5})^2 + 2(\sqrt{5})(2) + (2)^2 \over (\sqrt{5})^2 - (2)^2} \\ & = {5 + 4\sqrt{5} + 4 \over 5 - 4} \\ & = {9 + 4\sqrt{5} \over 1} \\ & = 9 + 4\sqrt{5} \end{align}

(ii)

\begin{align} {1 \over x} & = {1 \over 9 + 4\sqrt{5}} \\ & = {1 \over 9 + 4\sqrt{5}} \times {9 - 4\sqrt{5} \over 9 - 4\sqrt{5}} \\ & = {9 - 4\sqrt{5} \over (9 + 4\sqrt{5})(9 - 4\sqrt{5})} \\ & = {9 - 4\sqrt{5} \over (9)^2 - (4\sqrt{5})^2} \\ & = {9 - 4\sqrt{5} \over 81 - 80} \\ & = {9 - 4\sqrt{5} \over 1} \\ & = {9 - 4\sqrt{5}} \\ \\ \therefore x + {1 \over x} & = (9 + 4\sqrt{5}) + ({9 - 4\sqrt{5}}) \\ & = 9 + 4\sqrt{5} + 9 - 4\sqrt{5} \\ & = 18 \end{align}

(a)

\begin{align} 4\sqrt{2} - \sqrt{50} - {3 \over \sqrt{2}} + {7 \over \sqrt{72}} & = 4\sqrt{2} - \sqrt{25 \times 2} - {3 \over \sqrt{2}} + {7 \over \sqrt{36 \times 2}} \\ & = 4\sqrt{2} - \sqrt{25} \times \sqrt{2} - {3 \over \sqrt{2}} + {7 \over \sqrt{36} \times \sqrt{2}} \\ & = 4\sqrt{2} - 5\sqrt{2} - {3 \over \sqrt{2}} + {7 \over 6\sqrt{2}} \\ & = \sqrt{2} ( 4 - 5) - \left( {3 \over \sqrt{2}} - {7 \over 6\sqrt{2}} \right) \\ & = \sqrt{2} (-1) - \left( {3 \over \sqrt{2}} \times {6 \over 6} - {7 \over 6\sqrt{2}} \right) \\ & = -\sqrt{2} - \left( {18 \over 6\sqrt{2}} - {7 \over 6\sqrt{2}} \right) \\ & = -\sqrt{2} - \left( {18 - 7 \over 6\sqrt{2}} \right) \\ & = -\sqrt{2} - {11 \over 6\sqrt{2}} \\ & = -\sqrt{2} - {11 \over 6\sqrt{2}} \times {6\sqrt{2} \over 6\sqrt{2}} \phantom{00000000} [\text{Rationalise denominator}] \\ & = -\sqrt{2} - {11(6\sqrt{2}) \over (6\sqrt{2})^2} \\ & = -\sqrt{2} - {66\sqrt{2} \over 72} \\ & = -\sqrt{2} \left( 1 + {66 \over 72} \right) \\ & = -\sqrt{2} \left( {23 \over 12} \right) \\ & = -{23 \over 12}\sqrt{2} \end{align}

(b)

\begin{align} {\sqrt{12}\sqrt{8} \over 3} & = {\sqrt{4 \times 3} \sqrt{4 \times 2} \over 3} \\ & = {\sqrt{4} \times \sqrt{3} \times \sqrt{4} \times \sqrt{2} \over 3} \\ & = {2 \times \sqrt{3} \times 2 \times \sqrt{2} \over 3} \\ & = {4\sqrt{2}\sqrt{3} \over 3} \\ \\ {2 \over \sqrt{6}} & = {2 \over \sqrt{6}} \times {\sqrt{6} \over \sqrt{6}} \\ & = {2\sqrt{6} \over (\sqrt{6})^2} \\ & = {2\sqrt{6} \over 6} \\ & = {\sqrt{6} \over 3} \\ & = {\sqrt{2 \times 3} \over 3} \\ & = {\sqrt{2} \sqrt{3} \over 3} \\ \\ {3\sqrt{216} \over 2} & = {3\sqrt{36 \times 6} \over 2} \\ & = {3(\sqrt{36} \times \sqrt{6}) \over 2} \\ & = {3(6\sqrt{6}) \over 2} \\ & = {18\sqrt{6} \over 2} \\ & = 9\sqrt{6} \\ & = 9\sqrt{2 \times 3} \\ & = 9(\sqrt{2} \times \sqrt{3}) \\ & = 9\sqrt{2}\sqrt{3} \\ \\ \\ \therefore {1 \over \sqrt{2}} \times \left( {\sqrt{12}\sqrt{8} \over 3} - {2 \over \sqrt{6}} + {3\sqrt{216} \over 2} \right) & = k\sqrt{3} \\ {1 \over \sqrt{2}} \left( {4\sqrt{2}\sqrt{3} \over 3} - {\sqrt{2} \sqrt{3} \over 3} + 9\sqrt{2}\sqrt{3} \right) & = k\sqrt{3} \\ {4\sqrt{2}\sqrt{3} \over 3\sqrt{2}} - {\sqrt{2}\sqrt{3} \over 3\sqrt{2}} + {9\sqrt{2}\sqrt{3} \over \sqrt{2}} & = k\sqrt{3} \\ {4\sqrt{3} \over 3} - {\sqrt{3} \over 3} + 9\sqrt{3} & = k\sqrt{3} \\ {4 \over 3}\sqrt{3} - {1 \over 3}\sqrt{3} + 9\sqrt{3} & = k\sqrt{3} \\ \sqrt{3} \left( {4 \over 3} - {1 \over 3} + 9 \right) & = k\sqrt{3} \\ \sqrt{3} (10) & = k\sqrt{3} \\ 10\sqrt{3} & = k\sqrt{3} \\ \\ \therefore k & = 10 \end{align}

(a)

\begin{align} \sqrt{x^2 - 7} & = 3 \\ (\sqrt{x^2 - 7})^2 & = 3^2 \\ x^2 - 7 & = 9 \\ x^2 & = 9 + 7 \\ x^2 & = 16 \\ x & = \pm\sqrt{16} \\ & = \pm 4 \\ \text{For } & x = 4, \\ \text{L.H.S } & = \sqrt{(4)^2 - 7} \\ & = \sqrt{9} \\ & = 3 \\ & = \text{R.H.S} \\ \\ \text{For } & x = -4, \\ \text{L.H.S } & = \sqrt{(-4)^2 - 7} \\ & = \sqrt{9} \\ & = 3 \\ & = \text{R.H.S} \\ \\ \therefore x & = \pm 4 \end{align}

(b)

\begin{align} 2x + \sqrt{3 - 4x} & = 0 \\ \sqrt{3 - 4x} & = -2x \\ (\sqrt{3 - 4x})^2 & = (-2x)^2 \\ 3 - 4x & = 4x^2 \\ 0 & = 4x^2 + 4x - 3 \\ 0 & = (2x + 3)(2x - 1) \\ \\ 2x + 3 = 0 \phantom{00}&\text{or}\phantom{00} 2x - 1 = 0 \\ 2x = -3 \phantom{(} &\phantom{or00-1} 2x = 1 \\ x = -{3 \over 2} &\phantom{or00-12} x = {1 \over 2} \\ \\ \text{For } & x = -{3 \over 2}, \\ \text{L.H.S } & = 2\left(-{3 \over 2}\right) + \sqrt{3 - 4\left(-{3 \over 2}\right)} \\ & = -3 + \sqrt{9} \\ & = -3 + 3 \\ & = 0 \\ & = \text{R.H.S} \\ \\ \text{For } & x = {1 \over 2}, \\ \text{L.H.S } & = 2\left(1 \over 2\right) + \sqrt{3 - 4\left(1 \over 2\right)} \\ & = 1 + \sqrt{1} \\ & = 2 \\ & \ne \text{R.H.S} \\ \therefore \text{Reject } x & = {1 \over 2} \\ \\ \therefore x & = - {3 \over 2} \end{align}

(c)

\begin{align} {x \over \sqrt{1 - 8x}} & = {1 \over 3} \\ 3x & = \sqrt{1 - 8x} \\ (3x)^2 & = (\sqrt{1 - 8x})^2 \\ 9x^2 & = 1 - 8x \\ 0 & = -9x^2 - 8x + 1 \\ 0 & = 9x^2 + 8x - 1 \\ 0 & = (9x - 1)(x + 1) \\ \\ 9x - 1 = 0 \phantom{00}&\text{or}\phantom{00} x + 1 = 0 \\ 9x = 1 \phantom{00}&\phantom{or00+1} x = - 1 \\ x = {1 \over 9} \phantom{(.} & \\ \\ \text{For } & x = {1 \over 9}, \\ \text{L.H.S } & = {({1 \over 9}) \over \sqrt{1 - 8({1 \over 9})}} \\ & = {{1 \over 9} \over {1 \over 3}} \\ & = {1 \over 3} \\ & = \text{R.H.S} \\ \\ \text{For } & x = -1, \\ \text{L.H.S } & = {(-1) \over \sqrt{1 - 8(-1)}} \\ & = {-1 \over 3} \\ & = -{1 \over 3} \\ & \ne \text{R.H.S} \\ \therefore \text{Reject } x & = -1 \\ \\ \therefore x & = {1 \over 9} \end{align}

(d)

\begin{align} {4^x \over 8} & = \sqrt{2}(2^{x - 1}) \\ {4^x \over 8} & = (2^{1 \over 2})(2^{x - 1}) \\ {4^x \over 8} & = 2^{{1 \over 2} + (x - 1)} \phantom{00000000} [(a^m)(a^n) = a^{m + n} ] \\ {4^x \over 8} & = 2^{x - {1 \over 2}} \\ {(2^2)^x \over 2^3} & = 2^{x - {1 \over 2}} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {2^{2x} \over 2^3} & = 2^{x - {1 \over 2}} \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 2^{2x - 3} & = 2^{x - {1 \over 2}} \\ \\ \therefore 2x - 3 & = x - {1 \over 2} \\ 2x - x & = - {1 \over 2} + 3 \\ x & = {5 \over 2} \end{align}

\begin{align} 8 \times 4^y & = 2^{2x - 1} \\ 2^3 \times (2^2)^y & = 2^{2x - 1} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^3 \times 2^{2y} & = 2^{2x - 1} \\ [ a^m \times a^n = a^{m + n} ] \phantom{00000000} 2^{3 + 2y} & = 2^{2x - 1} \\ \\ \therefore 3 + 2y & = 2x - 1 \\ 2y & = 2x - 1 - 3 \\ 2y & = 2x - 4 \\ \\ y & = x - 2 \phantom{000} \text{ --- (1)} \end{align} \begin{align} 3^y\sqrt{3^x} & = 81 \\ 3^y \sqrt{3^x} & = 3^4 \\ [ \sqrt[n]{a} = a^{1 \over n} ] \phantom{00000000} 3^y (3^x)^{1 \over 2} & = 3^4 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 3^y (3^{{1 \over 2}x}) & = 3^4 \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 3^{y + {1 \over 2}x} & = 3^4 \\ \\ \therefore y + {1 \over 2}x & = 4 \\ 2y + x & = 8 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(x - 2) + x & = 8 \\ 2x - 4 + x & = 8 \\ 2x + x & = 8 + 4 \\ 3x & = 12 \\ x & = {12 \over 3} \\ & = 4 \\ \\ \text{Substitute } & x = 4 \text{ into (1),} \\ y & = (4) - 2 \\ & = 2 \\ \\ \therefore x & = 4, y = 2 \end{align}

(a)

\begin{align} 2(4^x) + 4^{x + 2} & = 9(4^{-0.5}) \\ 2(4^x) + 4^{x + 2} & = 9(0.5) \\ 2(4^x) + 4^{x + 2} & = 4.5 \\ [ a^{m + n} = (a^m)(a^n) ] \phantom{00000000} 2(4^x) + (4^x)(4^2) & = 4.5 \\ 2(4^x) + (4^x)(16) & = 4.5 \\ 2(4^x) + 16(4^x) & = 4.5 \\ \\ \text{When } & u = 4^x, \\ 2u + 16u & = 4.5 \\ 18u & = 4.5 \\ u & = {4.5 \over 18} \\ & = {1 \over 4} \\ \\ \text{Since } & u = 4^x, \\ 4^x & = {1 \over 4} \\ 4^x & = 4^{-1} \\ \\ \therefore x & = -1 \end{align}

(b)

\begin{align} 4^{x - 1} + 16^x & = 66 \\ 4^{x - 1} + (4^2)^x & = 66 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 4^{x - 1} + 4^{2x} & = 66 \\ 4^{x - 1} + (4^x)^2 & = 66 \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{00000000} {4^x \over 4^1} + (4^x)^2 & = 66 \\ {4^x \over 4} + (4^x)^2 & = 66 \\ \\ \text{When } & u = 4^x, \\ {u \over 4} + u^2 & = 66 \\ u + 4u^2 & = 264 \\ 4u^2 + u - 264 & = 0 \\ (4u + 33)(u - 8) & = 0 \end{align} \begin{align} 4u + 33 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 8 & = 0 \\ 4u & = -33 & u & = 8 \\ u & = -{33 \over 4} \\ \\ \text{Since } & u = 4^x, \\ 4^x & = -{33 \over 4} \text{ (N.A.)} & 4^x & = 8 \\ & & (2^2)^x & = 2^3 \\ & & 2^{2x} & = 2^3 \\ \\ & & \therefore 2x & = 3 \\ & & x & = {3 \over 2} \end{align}

(i)

\begin{align} \text{When } & t = 0, \\ m & = 300e^{-0.85(0)} \\ & = 300(1) \\ & = 300 \text{ mg} \end{align}

(ii)

\begin{align} \text{When } & t = 2, \\ m & = 300e^{-0.85(2)} \\ & = 54.80505 \\ & \approx 54.8 \text{ mg} \end{align}

(iii)

$$ \text{From part (i), vertical intercept is 300} $$

(Note time, t, cannot be negative, thus the graph starts from t = 0)

(iv)

\begin{align} \text{Amount of substance decayed} & = \text{Initial amount of substance} - \text{Amount of substance present} \\ p & = 300 - m \\ & = 300 - 300e^{-0.85t} \\ & = 300(1 - e^{-0.85t}) \end{align}

(a)

\begin{align} \sqrt{x - 8} \times \sqrt{x} & = 3 \\ \sqrt{(x - 8)(x)} & = 3 \\ \sqrt{x^2 - 8x} & = 3 \\ (\sqrt{x^2 - 8x})^2 & = 3^2 \\ x^2 - 8x & = 9 \\ x^2 - 8x - 9 & = 0 \\ (x - 9)(x + 1) & = 0 \\ \\ x - 9 = 0 \phantom{00}&\text{or}\phantom{00} x + 1 = 0 \\ x = 9 \phantom{00}&\phantom{or00+1} x = - 1 \\ \\ \text{For } & x = 9, \\ \text{L.H.S } & = \sqrt{9 - 8} \times \sqrt{9} \\ & = 1 \times 3 \\ & = 3 \\ & = \text{R.H.S} \\ \\ \text{For } & x = -1, \\ \text{L.H.S } & = \sqrt{-1 - 8} \times \sqrt{-1} \text{ (N.A.)} \\ \therefore \text{Reject } x & = - 1 \\ \\ \therefore x & = 9 \end{align}

(b)

\begin{align} \sqrt{18 - \sqrt{x - 1}} & = 4 \\ \left(\sqrt{18 - \sqrt{x - 1}} \right)^2 & = 4^2 \\ 18 - \sqrt{x - 1} & = 16 \\ -\sqrt{x - 1} & = -2 \\ \sqrt{x - 1} & = 2 \\ (\sqrt{x - 1})^2 & = 2^2 \\ x - 1 & = 4 \\ x & = 5 \\ \\ \text{For } & x = 5, \\ \text{L.H.S} & = \sqrt{18 - \sqrt{5 - 1}} \\ & = \sqrt{18 - \sqrt{4}} \\ & = \sqrt{18 - 2} \\ & = \sqrt{16} \\ & = 4 \\ & = \text{R.H.S} \\ \\ \therefore x & = 5 \end{align}

(c)

\begin{align} {\sqrt{11} + \sqrt{4} \over x} & = {x \over \sqrt{11} - \sqrt{4}} \\ {\sqrt{11} + 2 \over x} & = {x \over \sqrt{11} - 2} \\ (\sqrt{11} + 2)(\sqrt{11} - 2) & = x^2 \\ (\sqrt{11})^2 - (2)^2 & = x^2 \\ 11 - 4 & = x^2 \\ 7 & = x^2 \\ \\ x^2 & = 7 \\ x & = \pm\sqrt{7} \end{align}

(d)

\begin{align} 27(\sqrt{3})^x & = {3^x \over \sqrt[3]{9}} \\ (3^3)(\sqrt{3})^x & = {3^x \over \sqrt[3]{3^2}} \\ (3^3)(3^{1 \over 2})^x & = {3^x \over (3^2)^{1 \over 3}} \phantom{00000000} [ \sqrt[n]{a} = a^{1 \over n} ] \\ (3^3)(3^{{1 \over 2}x}) & = {3^x \over 3^{2 \over 3}} \phantom{0000000000} [ (a^m)^n = a^{mn} ] \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 3^{3 + {1 \over 2}x} & = {3^x \over 3^{2 \over 3}} \\ 3^{3 + {1 \over 2}x} & = 3^{x - {2 \over 3}} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \therefore 3 + {1 \over 2}x & = x - {2 \over 3} \\ {1 \over 2}x - x & = -{2 \over 3} - 3 \\ -{1 \over 2}x & = -{11 \over 3} \\ {1 \over 2}x & = {11 \over 3} \\ x & = {22 \over 3} \end{align}

\begin{align} (a - 6\sqrt{5})(2 + b\sqrt{5}) & = - 82 \\ (2)(a) + (a)(b\sqrt{5}) - (6\sqrt{5})(2) - (6\sqrt{5})(b\sqrt{5}) & = -82 \\ 2a + ab\sqrt{5} - 12\sqrt{5} - 30b & = -82 \\ 2a - 30b + ab\sqrt{5} - 12\sqrt{5} & = -82 \\ (2a - 30b) + (ab - 12)\sqrt{5} & = -82 \\ \\ \text{Since there are no terms with } \sqrt{5} & \text{ on the R.H.S,} \\ ab - 12 & = 0 \\ ab & = 12 \phantom{000} \text{ --- (1)} \\ \\ \text{Comparing the non-surd terms,} & \\ 2a - 30b & = -82 \\ 2a & = 30b - 82 \\ a & = 15b - 41 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ (15b - 41) b & = 12 \\ 15b^2 - 41b - 12 & = 0 \\ (15b + 4)(b - 3) & = 0 \end{align} \begin{align} 15b + 4 & = 0 \phantom{00}&\text{or}\phantom{0000} b - 3 & = 0 \\ 15b & = -4 & b & = 3 \\ b & = -{4 \over 15} \\ \\ \text{Substitute } & \text{into (2),} & \text{Substitute } & \text{into (2),} \\ a & = 15 \left(-{4 \over 15}\right) - 41 &\text{or}\phantom{0000} a & = 15(3) - 41 \\ & = -45 & & = 4 \end{align}

(i)

\begin{align} \text{Half the length of } BC & = {8\sqrt{3} - 2\sqrt{2} \over 2} \\ & = {8\sqrt{3} \over 2} - {2\sqrt{2} \over 2} \\ & = (4\sqrt{3} - \sqrt{2}) \text{ cm} \\ \\ \text{Area } & = {1 \over 2} \times \text{Base } \times {Height} \\ 46 & = {1 \over 2} \times BC \times h \\ 92 & = BC \times h \\ \\ h & = {92 \over BC} \\ & = {92 \over 8\sqrt{3} - 2\sqrt{2}} \\ & = {92 \over 8\sqrt{3} - 2\sqrt{2}} \times {8\sqrt{3} + 2\sqrt{2} \over 8\sqrt{3} + 2\sqrt{2}} \phantom{00000000} [\text{Rationalise denominator}] \\ & = {92(8\sqrt{3} + 2\sqrt{2}) \over (8\sqrt{3} - 2\sqrt{2})(8\sqrt{3} + 2\sqrt{2})} \\ & = {736\sqrt{3} + 184\sqrt{2} \over (8\sqrt{3})^2 - (2\sqrt{2})^2} \\ & = {736\sqrt{3} + 184\sqrt{2} \over 192 - 8} \\ & = {736\sqrt{3} + 184\sqrt{2} \over 184} \\ & = {736\sqrt{3} \over 184} + {184\sqrt{2} \over 184} \\ & = (4\sqrt{3} + \sqrt{2}) \text{ cm} \end{align}

(ii)

\begin{align} \text{By } & \text{Pythagoras theorem,} \\ AB^2 & = (4\sqrt{3} - \sqrt{2})^2 + (4\sqrt{3} + \sqrt{2})^2 \\ & = [(4\sqrt{3})^2 - 2(4\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2] + [(4\sqrt{3})^2 + 2(4\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2] \\ & = [48 - 8\sqrt{3}\sqrt{2} + 2] + [48 + 8\sqrt{3}\sqrt{2} + 2] \\ & = [50 - 8\sqrt{3}\sqrt{2}] + [50 + 8\sqrt{3}\sqrt{2}] \\ & = 50 - 8\sqrt{3}\sqrt{2} + 50 + 8\sqrt{3}\sqrt{2} \\ & = 100 \\ \\ AB & = \pm\sqrt{100} \\ & = \pm 10 \\ & = 10 \text{ or } - 10 \text{ (Reject)} \\ \\ \text{Perimeter } & = AB + BC + AC \\ & = 10 + (8\sqrt{3} - 2\sqrt{2}) + 10 \\ & = 10 + 8\sqrt{3} - 2\sqrt{2} + 10 \\ & = (20 + 8\sqrt{3} - 2\sqrt{2}) \text{ cm} \end{align}

\begin{align} y & = ax^n - 23 \\ \\ \text{Using the point } (3, 4), & \text{ substitute } x = 3 \text{ and } y = 4, \\ 4 & = a(3^n) -23 \\ 27 & = a(3^n) \\ {27 \over 3^n} & = a \\ {3^3 \over 3^n} & = a \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 3^{3 - n} & = a \phantom{000} \text{ --- (1)} \\ \\ \text{Using the point } (9, 200), & \text{ substitute } x = 9 \text{ and } y = 220, \\ 220 & = a(9^n) - 23 \\ 220 + 23 & = a(9^n) \\ 243 & = a(9^n) \\ {243 \over 9^n} & = a \\ {3^5 \over (3^2)^n} & = a \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {3^5 \over 3^2n} & = a \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 3^{5 - 2n} & = a \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3^{5 - 2n} & = 3^{3 - n} \\ \\ \therefore 5 - 2n & = 3 - n \\ -2n + n & = 3 - 5 \\ -n & = -2 \\ n & = 2 \\ \\ \text{Substitute } & n = 2 \text{ into (1),} \\ 3^{5 - 2(2)} & = a \\ 3^{1} & = a \\ 3 & = a \\ \\ \text{When } n = 2 & \text{ and } a = 3, \\ y & = ax^n - 23 \\ & = 3x^2 - 23 \\ \\ \text{Using the point } (-1, k), & \text{ substitute } x = -1 \text{ and } y = k, \\ k & = 3(-1)^2 - 23 \\ & = -20 \\ \\ \therefore a & = 3, n = 2, k = -20 \end{align}

(a)

\begin{align} 5^x + 5 & = 30(5^{x - 1}) \\ 5^x + 5 & = 30 \left( 5^x \over 5^1 \right) \phantom{00000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ 5^x + 5 & = 30 \left( 5^x \over 5 \right) \\ 5^x + 5 & = 6(5^x) \\ \\ \text{Let } & u = 5^x, \\ u + 5 & = 6u \\ u - 6u & = - 5 \\ -5u & = -5 \\ 5u & = 5 \\ u & = {5 \over 5} \\ & = 1 \\ \\ \text{Since } & u = 5^x, \\ 5^x & = 1 \\ 5^x & = 5^0 \\ \\ \therefore x & = 0 \end{align}

(b)

\begin{align} 9^x + 10(3^x) & = 3^{x + 2} + 12 \\ (3^2)^x + 10(3^x) & = 3^{x + 2} + 12 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 3^{2x} + 10(3^x) & = 3^{x + 2} + 12 \\ (3^x)^2 + 10(3^x) & = 3^{x + 2} + 12 \\ (3^x)^2 + 10(3^x) & = (3^x)(3^2) + 12 \phantom{00000000} [ a^{m + n} = (a^m)(a^n) ] \\ (3^x)^2 + 10(3^x) & = (3^x)(9) + 12 \\ (3^x)^2 + 10(3^x) & = 9(3^x) + 12 \\ \\ \text{Let } & u = 3^x, \\ u^2 + 10u & = 9u + 12 \\ u^2 + 10u - 9u - 12 & = 0 \\ u^2 + u - 12 & = 0 \\ (u + 4)(u - 3) & = 0 \end{align} \begin{align} u + 4 & = 0 \phantom{00} & \text{or}\phantom{0000} u - 3 & = 0 \\ u & = -4 & u & = 3 \\ \\ \text{Since } & u = 3^x, & \text{Since } & u = 3^x, \\ 3^x & = -4 \text{ (N.A.)} & 3^x & = 3 \\ & & 3^x & = 3^1 \\ \\ & & \therefore x & = 1 \end{align}

(i)

\begin{align} \text{When } & t = 0, \\ \theta & = 20 + 100(0.8)^{0 \over 6} \\ & = 20 + 100(1) \\ & = 120 \end{align}

(ii)

\begin{align} \text{When } & t = 8, \\ \theta & = 20 + 100(0.8)^{8 \over 6} \\ & = 94.265 \phantom{.} 42 \\ & \approx 94.3 \end{align}

(iii)

\begin{align} \text{When } & \theta = 84, \\ 84 & = 20 + 100(0.8)^{t \over 6} \\ 64 & = 100(0.8)^{t \over 6} \\ {64 \over 100} & = (0.8)^{t \over 6} \\ 0.64 & = (0.8)^{t \over 6} \\ (0.8)^2 & = (0.8)^{t \over 6} \\ \\ \therefore 2 & = {t \over 6} \\ 6(2) & = t \\ 12 & = t \end{align}

(iv)

\begin{align} \theta & = 20 + 100(0.8)^{t \over 6} \end{align} \begin{align} \text{As } t \rightarrow \infty &, \phantom{0} (0.8)^{t \over 6} \rightarrow 0 \\ \\ \therefore \theta & = 20 + 100(0) \\ & = 20 \end{align}