(a)

\begin{align} \log_x 27 & = 1.5 \\ \log_x 27 & = {3 \over 2} \\ \\ \implies 27 & = x^{3 \over 2} \phantom{0000000000} [\text{Change to index form}] \\ (27)^2 & = (x^{3 \over 2})^2 \\ 729 & = x^3 \\ \sqrt[3] {729} & = x \\ 9 & = x \end{align}

(b)

\begin{align} \log_2 x \log_8 x & = 12 \\ [\text{Change of base}] \phantom{00000000} \log_2 x \left( \log_2 x \over \log_2 8 \right) & = 12 \\ \left(\log_2 x \over 1\right) \left( \log_2 x \over \log_2 8 \right) & = 12 \\ {(\log_2 x)^2 \over \log_2 8 } & = 12 \\ {(\log_2 x)^2 \over \log_2 2^3} & = 12 \\ [\text{Power law}] \phantom{0000000000000.} {(\log_2 x)^2 \over 3\log_2 2} & = 12 \\ {(\log_2 x)^2 \over 3(1)} & = 12 \\ {(\log_2 x)^2 \over 3} & = 12 \\ (\log_2 x)^2 & = 3(12) \\ (\log_2 x)^2 & = 36 \\ \log_2 x & = \pm \sqrt{36} \\ \log_2 x & = \pm 6 \\ \\ \implies x & = 2^{\pm 6} \phantom{0000000000} [\text{Change to index form}] \\ & = 2^6 \text{ or } 2^{-6} \\ & = 64 \text{ or } {1 \over 64} \end{align}

(c)

\begin{align} \log_3 (x - 2) & = 3 - \log_3 (x + 4) \\ \log_3 (x - 2) + \log_3 (x + 4) & = 3 \\ [\text{Product law}] \phantom{00000000} \log_3 [(x - 2)(x + 4)] & = 3 \\ \log_3 (x^2 + 4x - 2x - 8) & = 3 \\ \log_3 (x^2 + 2x - 8) & = 3 \\ \\ \implies x^2 + 2x - 8 & = 3^3 \phantom{00000000} [\text{Change to index form}] \\ x^2 + 2x - 8 & = 27 \\ x^2 + 2x - 35 & = 0 \\ (x - 5)(x + 7) & = 0 \end{align}
\begin{align} x - 5 & = 0 \phantom{00}&\text{or}\phantom{0000} x + 7 & = 0 \\ x & = 5 & x & = -7 \text{(Reject, since } \log_3 (-9) \text{ and } \log_3 (-3) \text{ are undefined}) \end{align}

(a)

\begin{align} \require{cancel} \log_3 xy - \log_3 (x - 1) & = \log_3 6x^2 - 1 \\ [\text{Quotient law}] \phantom{000000000000} \log_3 \left( {xy \over x - 1} \right) & = \log_3 6x^2 - 1 \\ \log_3 \left( {xy \over x - 1} \right) - \log_3 6x^2 & = -1 \\ \log_3 \left( {{xy \over x - 1} \over 6x^2} \right) & = -1 \\ \log_3 \left( {xy \over x - 1} \div 6x^2 \right) & = -1 \\ \log_3 \left( {\cancel{x}y \over x - 1} \times {1 \over 6x^\cancel{2}} \right) & = -1 \\ \log_3 \left[ {y \over 6x(x - 1)} \right] & = -1 \\ \\ \implies {y \over 6x(x - 1)} & = 3^{-1} \phantom{0000000000} [\text{Change to index form}] \\ {y \over 6x(x - 1)} & = {1 \over 3} \\ 3y & = 6x(x - 1) \\ y & = 2x(x - 1) \\ y & = 2x^2 - 2x \end{align}

(b)

\begin{align} 2\log_5 x & = 5 - \log_x 25 \\ 2\log_5 x & = 5 - {\log_5 25 \over \log_5 x} \phantom{00000000} [\text{Change of base}] \\ 2\log_5 x & = 5 - {\log_5 5^2 \over \log_5 x} \\ 2\log_5 x & = 5 - {2\log_5 5 \over \log_5 x} \phantom{00000000} [\text{Power law}] \\ 2\log_5 x & = 5 - {2(1) \over \log_5 x} \\ 2\log_5 x & = 5 - {2 \over \log_5 x} \\ \\ \text{Let } & u = \log_5 x, \\ 2u & = 5 - {2 \over u} \\ 2u^2 & = 5u - 2 \\ 2u^2 - 5u + 2 & = 0 \\ (2u - 1)(u - 2) & = 0 \end{align} \begin{align} 2u - 1 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 2 & = 0 \\ 2u & = 1 \\ u & = {1 \over 2} & u & = 2 \\ \\ \text{Since } & u = \log_5 x, & \text{Since } & u = \log_5 x, \\ \log_5x & = {1 \over 2} & \log_5 x & = 2 \\ \\ \implies x & = 5^{1 \over 2} & x & = 5^2 \\ & = \sqrt{5} & & = 25 \end{align}

(a)

\begin{align} \log_3 (x - 19) & = 4 \\ \\ \implies x - 19 & = 3^4 \phantom{00000000} [\text{Change to index form}] \\ x - 19 & = 81 \\ x & = 81 + 19 \\ x & = 100 \\ \\ \therefore \lg x & = \lg 100 \\ & = 2 \end{align}

(b)

\begin{align} \lg (x + 2) + 7\lg 2 & = \log_3 9 + \lg (2x + 1) \\ [\text{Power law}] \phantom{00000000} \lg (x + 2) + \lg 2^7 & = \log_3 9 + \lg (2x + 1) \\ \lg (x + 2) + \lg 128 & = \log_3 9 + \lg (2x + 1) \\ \lg (x + 2) + \lg 128 & = \log_3 3^2 + \lg (2x + 1) \\ \lg (x + 2) + \lg 128 & = 2\log_3 3 + \lg (2x + 1) \\ \lg (x + 2) + \lg 128 & = 2(1) + \lg (2x + 1) \\ \lg (x + 2) + \lg 128 & = 2 + \lg (2x + 1) \\ [\text{Product law}] \phantom{000000000} \lg [128(x + 2)] & = 2 + \lg (2x + 1) \\ \lg (128x + 256) & = 2 + \lg (2x + 1) \\ \lg (128x + 256) - \lg (2x + 1) & = 2 \\ [\text{Quotient law}] \phantom{000000} \lg \left(128x + 256 \over 2x + 1\right) & = 2 \\ \log_{10} \left(128x + 256 \over 2x + 1\right) & = 2 \\ \\ \implies {128x + 256 \over 2x + 1} & = 10^2 \phantom{0000000000} [\text{Change to index form}] \\ {128x + 256 \over 2x + 1} & = 100 \\ 128x + 256 & = 100(2x + 1) \\ 128x + 256 & = 200x + 100 \\ 128x - 200x & = 100 - 256 \\ -72x & = -156 \\ x & = {-156 \over -72} \\ & = 2{1 \over 6} \end{align}

(c)

\begin{align} e^{2x} + e^x - 6 & = 0 \\ (e^x)^2 + e^x - 6 & = 0 \\ \\ \text{Let } & u = e^x, \\ u^2 + u - 6 & = 0 \\ (u + 3)(u - 2) & = 0 \end{align} \begin{align} u + 3 & = 0 \phantom{00} & \text{or} \phantom{0000} u - 2 & = 0 \\ u & = -3 & u & = 2 \\ \\ \text{Since } & u = e^x, & \text{Since } & u = e^x, \\ e^x & = -3 \text{ (Reject, } e^x > 0) & e^x & = 2 \\ & & \ln e^x & = \ln 2 \\ & & x\ln e & = \ln 2 \\ & & x(1) & = \ln 2 \\ & & x & = \ln 2 \\ & & & = 0.69314 \\ & & & \approx 0.693 \end{align}

(i)

\begin{align} \log_2 x & = a \\ \\ \implies x & = 2^a \\ \\ \log_8 y & = b \\ \\ \implies y & = 8^b \\ & = (2^3)^b \\ & = 2^{3b} \end{align}
\begin{align} x^2y & = (2^a)^2 (2^{3b}) \\ & = (2^{2a})(2^{3b}) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = 2^{2a + 3b} \phantom{00000000000} [ (a^m)(a^n) = a^{m + n} ] \end{align} \begin{align} {x \over y} & = {2^a \over 2^{3b}} \\ & = 2^{a - 3b} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \end{align}

(ii)

\begin{align} x^2y & = 2^{2a + 3b} \\ 32 & = 2^{2a + 3b} \\ 2^5 & = 2^{2a + 3b} \\ \\ \therefore 5 & = 2a + 3b \phantom{0} \text{ --- (1)} \\ \\ \\ {x \over y} & = 2^{a - 3b} \\ 0.5 & = 2^{a - 3b} \\ {1 \over 2} & = 2^{a - 3b} \\ 2^{-1} & = 2^{a - 3b} \\ \\ \therefore -1 & = a - 3b \\ \\ a & = 3b - 1 \phantom{0} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 5 & = 2(3b - 1) + 3b \\ 5 & = 6b - 2 + 3b \\ 7 & = 9b \\ {7 \over 9} & = b \\ \\ \text{Substitute } & b = {7 \over 9} \text{ into (2),} \\ a & = 3\left(7 \over 9\right) - 1 \\ & = {7 \over 3} - 1 \\ & = {4 \over 3} \end{align}

(i) 15 years has elapsed since the beginning of 1990 to the beginning of 2005

\begin{align} \text{Let } P \text{ represent } & \text{the population of the town.} \\ \\ P & = 12000e^{0.07n} \\ \\ \text{When } & n = 15, \\ P & = 12000e^{0.07(15)} \\ & = 34291.81 \\ & \approx 34292 \end{align}

(ii)

\begin{align} \text{When } & P = 90000, \\ 90000 & = 12000e^{0.07n} \\ {90000 \over 12000} & = e^{0.07n} \\ 7.5 & = e^{0.07n} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln 7.5 & = \ln e^{0.07n} \phantom{00000000} [\text{Power law}] \\ \ln 7.5 & = 0.07n \ln e \\ \ln 7.5 & = 0.07n (1) \\ \ln 7.5 & = 0.07n \\ \\ n & = {\ln 7.5 \over 0.07} \\ & = 28.784 \\ \\ \text{Year} & = 1990 + 28 \\ & = 2018 \end{align}

\begin{align} \log_b (xy^2) & = m \\ [\text{Product law}] \phantom{00000000} \log_b x + \log_b y^2 & = m \\ [\text{Power law}] \phantom{0000000.} \log_b x + 2\log_b y & = m \\ 2\log_b y & = m - \log_b x \\ \log_b y & = {1 \over 2}m - {1 \over 2}\log_b x \phantom{0} \text{ --- (1)} \end{align} \begin{align} \log_b (x^3y) & = n \\ \log_b x^3 + \log_b y & = n \\ 3\log_b x + \log_b y & = n \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3\log_b x + \left({1 \over 2}m - {1 \over 2}\log_b x\right) & = n \\ 3\log_b x + {1 \over 2}m - {1 \over 2} \log_b x & = n \\ 3\log_b x - {1 \over 2} \log_b x & = n - {1 \over 2}m \\ (\log_b x) \left(3 - {1 \over 2}\right) & = n - {1 \over 2}m \\ (\log_b x) \left(5 \over 2\right) & = n - {1 \over 2}m \\ {5 \over 2}\log_b x & = n - {1 \over 2}m \\ 5\log_b x & = 2n - m \\ \log_b x & = {2n - m \over 5} \phantom{000} \text{ --- (3)} \\ \\ \text{Substitute } & \log_b x = {2n - m \over 5} \text{ into (1),} \\ \log_b y & = {1 \over 2}m - {1 \over 2} \left(2n - m \over 5 \right) \\ & = {m \over 2} - {2n - m \over 10} \\ & = {5m \over 10} - {2n - m \over 10} \\ & = {5m - (2n - m) \over 10} \\ & = {5m - 2n + m \over 10} \\ & = {6m - 2n \over 10} \\ & = {2(3m - n) \over 10} \\ \log_b y & = {3m - n \over 5} \phantom{000} \text{ --- (4)} \\ \end{align}
\begin{align} \therefore \log_b {y \over x} & = \log_b y - \log_b x \phantom{00000000} [\text{Quotient law}] \\ & = {3m - n \over 5} - {2n - m \over 5} \phantom{000.0} [\text{Use (3) & (4)}] \\ & = {3m - n - (2n - m) \over 5} \\ & = {3m - n - 2n + m \over 5} \\ & = {4m - 3n \over 5} \\ \\ \\ \therefore \log_b \sqrt{xy} & = \log_b (xy)^{1 \over 2} \\ & = {1 \over 2} \log_b xy \phantom{00000000000000.} [\text{Power law}] \\ & = {1 \over 2} (\log_b x + \log_b y) \phantom{0000000.} [\text{Product law}] \\ & = {1 \over 2} \left( {2n - m \over 5} + {3m - n \over 5} \right) \phantom{00} [\text{Use (3) & (4)}] \\ & = {1 \over 2} \left( {2n - m + 3m - n \over 5} \right) \\ & = {1 \over 2} \left( {n + 2m \over 5} \right) \\ & = {1 \over 10} (n + 2m) \end{align}

(i)(a)

\begin{align} \text{When } & x = 4, \\ y & = 2\lg 4 \\ & = 1.2041 \\ & \approx 1.20 \end{align}

(i)(b)

\begin{align} \text{When } & y = 1.5, \\ 1.5 & = 2\lg x \\ 1.5 \div 2 & = \lg x \\ 0.75 & = \log_{10} x \\ \\ \implies 10^{0.75} & = x \phantom{0000000000} [\text{Change to index form}] \\ 5.6234 & = x \\ 5.62 & \approx x \end{align}

(ii)

\begin{align} \text{When } & y = 0, \\ 0 & = 2\lg x \\ 0 & = \lg x \\ 0 & = \log_{10} x \\ \\ \implies 10^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ \implies & x\text{-intercept is } 1 \end{align}

(iii)

\begin{align} x\sqrt{10} & = \sqrt[4]{10^x} \\ x(10)^{1 \over 2} & = (10^x)^{1 \over 4} \\ x(10)^{1 \over 2} & = 10^{{1 \over 4}x} \\ x & = {10^{{1 \over 4}x} \over 10^{1 \over 2}} \\ x & = 10^{{1 \over 4}x - {1 \over 2}} \phantom{0000000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg x & = \lg 10^{{1 \over 4}x - {1 \over 2}} \\ \lg x & = \left({1 \over 4}x - {1 \over 2}\right)(\lg 10) \phantom{00000} [\text{Power law}] \\ \lg x & = \left({1 \over 4}x - {1 \over 2}\right) (1) \\ \lg x & = {1 \over 4}x - {1 \over 2} \\ 2\lg x & = {1 \over 2}x - 1 \\ \\ \text{Since the } & \text{graph is } y = 2\lg x, \\ \therefore \text{Draw } & y = {1 \over 2}x - 1 \end{align}

(a)

\begin{align} \log_9 (3^{x + 1}) & = x^2 \\ [\text{Power law}] \phantom{00000000} (x + 1) \log_9 3 & = x^2 \\ (x + 1) \log_9 \sqrt{9} & = x^2 \\ (x + 1) \log_9 9^{1 \over 2} & = x^2 \\ (x + 1)\left(1 \over 2\right)\log_9 9 & = x^2 \\ (x + 1)\left(1 \over 2\right)(1) & = x^2 \\ {1 \over 2}(x + 1) & = x^2 \\ x + 1 & = 2x^2 \\ 0 & = 2x^2 - x - 1 \\ 0 & = (2x + 1)(x - 1) \\ \\ 2x + 1 = 0 \phantom{0000} & \text{or}\phantom{0000} x - 1 = 0 \\ 2x = -1 \phantom{00.} & \phantom{or0000-1} x = 1 \\ x = -{1 \over 2} \phantom{0(} & \end{align}

(b)

\begin{align} 4^{3x} + \log_2 \left(1 \over 8\right) & = 5 \\ 4^{3x} + \log_2 \left(1 \over 2^3 \right) & = 5 \\ 4^{3x} + \log_2 2^{-3} & = 5 \\ [\text{Power law}] \phantom{00000000} 4^{3x} + (-3)\log_2 2 & = 5 \\ 4^{3x} + (-3)(1) & = 5 \\ 4^{3x} - 3 & = 5 \\ 4^{3x} & = 5 + 3 \\ 4^{3x} & = 8 \\ (2^2)^{3x} & = 2^3 \\ [ (a^m)^n = a^{mn}] \phantom{00000000000000000} 2^{2(3x)} & = 2^3 \\ 2^{6x} & = 2^3 \\ \\ \therefore 6x & = 3 \\ x & = {3 \over 6} \\ & = {1 \over 2} \end{align}

(c)

\begin{align} \log_2 x^2 - \log_2 (2x + 5) & = 2 \\ [\text{Quotient law}] \phantom{00000000} \log_2 \left({x^2 \over 2x + 5}\right) & = 2 \\ \\ \implies {x^2 \over 2x + 5} & = 2^2 \phantom{00000000} [\text{Change to index form}] \\ {x^2 \over 2x + 5} & = 4 \\ x^2 & = 4(2x + 5) \\ x^2 & = 8x + 20 \\ x^2 - 8x - 20 & = 0 \\ (x - 10)(x + 2) & = 0 \\ \\ x - 10 = 0 \phantom{0000} & \text{or}\phantom{0000} x + 2 = 0 \\ x = 10 \phantom{000} & \phantom{or0000+2} x = - 2 \end{align}

(d)

\begin{align} \log_2 x & = 4\log_x 2 \\ \log_2 x & = 4 \left( \log_2 2 \over \log_2 x \right) \phantom{00000000} [\text{Change of base}] \\ \log_2 x & = 4 \left( 1 \over \log_2 x \right) \\ \log_2 x & = {4 \over \log_2 x } \\ (\log_2 x)^2 & = 4 \\ \log_2 x & = \pm \sqrt{4} \\ \log_2 x & = \pm 2 \\ \\ \implies x & = 2^{\pm 2} \phantom{0000000000000000} [\text{Change to index form}] \\ x & = 2^2 \text{ or } 2^{-2} \\ x & = 4 \text{ or } {1 \over 4} \end{align}

(i)

\begin{align} \log_2 x & = p \phantom{0} \text{ --- (1)} \\ \\ \log_4 y & = q \phantom{0} \text{ --- (2)} \end{align} \begin{align} \text{From (1), } \log_2 x & = p \\ \\ \implies x & = 2^p \phantom{0} \text{ --- (3)} \\ \\ \text{From (2), } \log_4 y & = q \\ \\ \implies y & = 4^q \\ & = (2^2)^q \\ & = 2^{2q} \phantom{0} \text{ --- (4)} \end{align}
\begin{align} \log_2 xy & = \log_2 x + \log_2 y \phantom{00000000} [\text{Product law}] \\ & = p + \log_2 2^{2q} \phantom{0000000000} [\text{Use (1) & (4)}] \\ & = p + 2q \log_2 2 \phantom{000000000.} [\text{Power law}] \\ & = p + 2q (1) \\ & = p + 2q \end{align}

(ii)

\begin{align} \log_4 {x \over y} & = {\log_2 {x \over y} \over \log_2 4} \phantom{000000000000} [\text{Change of base}] \\ & = {\log_2 {x \over y} \over \log_2 2^2} \\ & = {\log_2 {x \over y} \over 2\log_2 2} \phantom{00000000000.} [\text{Power law}] \\ & = {\log_2 {x \over y} \over 2(1)} \\ & = {\log_2 {x \over y} \over 2} \\ & = {1 \over 2}\log_2 {x \over y} \\ & = {1 \over 2} (\log_2 x - \log_2 y) \phantom{0000} [\text{Quotient law}] \\ & = {1 \over 2} (p - \log_2 2^{2q}) \phantom{000000} [\text{Use (1) & (4)}] \\ & = {1 \over 2} [p - (2q)\log_2 2] \\ & = {1 \over 2} [p - (2q)(1)] \\ & = {1 \over 2}(p - 2q) \\ & = {1 \over 2}p - q \end{align}

(iii)

\begin{align} \log_x 4y & = {\log_2 4y \over \log_2 x} \phantom{0000000000} [\text{Change of base}] \\ & = {\log_2 4 + \log_2 y \over \log_2 x} \phantom{0000} [\text{Product law}] \\ & = {\log_2 2^2 + \log_2 y \over \log_2 x} \\ & = {2\log_2 2 + \log_2 y \over \log_2 x} \phantom{000} [\text{Power law}] \\ & = {2(1) + \log_2 y \over \log_2 x} \\ & = {2 + \log_2 y \over \log_2 x} \\ & = {2 + \log_2 2^{2q} \over p} \phantom{0000000} [\text{Use (1) & (4)}] \\ & = {2 + (2q)\log_2 2 \over p} \\ & = {2 + (2q)(1) \over p} \\ & = {2 + 2q \over p} \end{align}

(iv)

\begin{align} x^2y & = (2^p)^2 (2^{2q}) \phantom{00000000} [\text{Use (3) & (4)}] \\ & = (2^{2p})(2^{2q}) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = 2^{2p + 2q} \phantom{00000000000} [ (a^m)(a^n) = a^{m + n}] \end{align}

(a)

\begin{align} 2^x & = 128(4^y) \\ 2^x & = (2^7)[(2^2)^y)] \\ 2^x & = (2^7)(2^{2y}) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ 2^x & = 2^{7 + 2y} \phantom{0000000000(} [ (a^m)(a^n) = a^{m + n}] \\ \\ \therefore x & = 7 + 2y \phantom{0} \text{ --- (1)} \\ \\ \\ \ln (4x + y) & = \ln 40 - 2\ln 2 \\ \ln (4x + y) & = \ln 40 - \ln 2^2 \phantom{00000000} [\text{Power law}] \\ \ln (4x + y) & = \ln 40 - \ln 4 \\ \ln (4x + y) & = \ln \left(40 \over 4\right) \phantom{0000000000(} [\text{Quotient law}] \\ \ln (4x + y) & = \ln 10 \\ \\ \therefore 4x + y & = 10 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 4(7 + 2y) + y & = 10 \\ 28 + 8y + y & = 10 \\ 9y & = -18 \\ y & = -2 \\ \\ \text{Substitute } & y = - 2 \text{ into (1),} \\ x & = 7 + 2(-2) \\ & = 3 \end{align}

(b)(i)

\begin{align} \lg (1 + 2x) - \lg x^2 & = 1 - \lg (2 + 5x) \\ [\text{Quotient law}] \phantom{00000000000000} \lg \left(1 + 2x \over x^2\right) & = 1 - \lg (2 + 5x) \\ \lg \left(1 + 2x \over x^2\right) + \lg (2 + 5x) & = 1 \\ [\text{Product law}] \phantom{000000} \lg \left[ (1 + 2x)(2 + 5x) \over x^2 \right] & = 1 \\ \log_{10} \left[ (1 + 2x)(2 + 5x) \over x^2 \right] & = 1 \\ \\ \implies { (1 + 2x)(2 + 5x) \over x^2 } & = 10^1 \\ { (1 + 2x)(2 + 5x) \over x^2} & = 10 \\ (1 + 2x)(2 + 5x) & = 10x^2 \\ 2 + 5x + 4x + 10x^2 & = 10x^2 \\ 9x & = - 2 \\ x & = -{2 \over 9} \end{align}

(b)(ii)

\begin{align} 3^{y + 2} & = 5^y \\ [ a^{m + n} = a^m \times a^n ] \phantom{00000000} 3^y \times 3^2 & = 5^y \\ 3^y \times 9 & = 5^y \\ 9(3^y) & = 5^y \\ 9 & = {5^y \over 3^y} \\ 9 & = \left(5 \over 3\right)^y \phantom{00000000} \left[ {a^m \over b^m} = \left(a \over b\right)^m \right] \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 9 & = \lg \left(5 \over 3\right)^y \\ \lg 9 & = y \lg {5 \over 3} \phantom{000000000} [\text{Power law}] \\ {\lg 9 \over \lg {5 \over 3}} & = y \\ 4.3013 & = y \\ 4.30 & \approx y \end{align}

(i)

\begin{align} \ln \sqrt[3]{10e} & = \ln (10e)^{1 \over 3} \\ & = {1 \over 3} \ln (10e) \phantom{0000000000} [\text{Power law}] \\ & = {1 \over 3} (\ln 10 + \ln e) \phantom{00000(} [\text{Product law}] \\ & = {1 \over 3} (\ln 10 + 1) \\ & = {1 \over 3} [\ln (5 \times 2) + 1 ] \\ & = {1 \over 3}( \ln 5 + \ln 2 + 1) \\ & = {1 \over 3}(b + a + 1) \end{align}

(ii)

\begin{align} \ln x & = {b - 2a \over 2} \\ 2\ln x & = b - 2a \\ 2\ln x & = \ln 5 - 2\ln 2 \\ \ln x^2 & = \ln 5 - \ln 2^2 \phantom{00000000} [\text{Power law}] \\ \ln x^2 & = \ln 5 - \ln 4 \\ \ln x^2 & = \ln \left(5 \over 4\right) \phantom{0000000000(} [\text{Quotient law}] \\ \\ \therefore x^2 & = {5 \over 4} \\ x & = \pm \sqrt{5 \over 4} \\ & = \pm {\sqrt{5} \over 2} \\ & = {\sqrt{5} \over 2} \text{ or } - {\sqrt{5} \over 2} \phantom{0} \left( \text{Reject, since } \ln -{\sqrt{5} \over 2} \text{ is undefined} \right) \end{align}

\begin{align} y & = \ln (2x + e^2) \\ \\ \text{When } x = 0 & \text{ and } y = h, \\ h & = \ln [2(0) + e^2] \\ h & = \ln e^2 \\ h & = 2\ln e \phantom{00000000} [\text{Power law}] \\ & = 2(1) \\ & = 2 \end{align} \begin{align} \text{When } x = k & \text{ and } y = 0, \\ 0 & = \ln [2(k) + e^2] \\ 0 & = \ln (2k + e^2) \\ 0 & = \log_e (2k + e^2) \\ & \downarrow \\ e^0 & = 2k + e^2 \phantom{00000000} [\text{Change to index form}] \\ 1 & = 2k + e^2 \\ 1 - e^2 & = 2k \\ \\ \therefore k & = {1 - e^2 \over 2} \end{align}
\begin{align} y & = {1 \over 2} - e^{ax} \\ \\ \text{When } x = h = 2 & \text{ and } y = k = {1 - e^2 \over 2}, \\ {1 - e^2 \over 2} & = {1 \over 2} - e^{a(2)} \\ 1 - e^2 & = 1 - 2e^{2a} \\ -e^2 & = - 2e^{2a} \\ e^2 & = 2e^{2a} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln e^2 & = \ln (2e^{2a}) \\ \ln e^2 & = \ln 2 + \ln e^{2a} \phantom{00000000} [\text{Product law}] \\ (2)\ln e & = \ln 2 + (2a)\ln e \phantom{000000} [\text{Power law}] \\ (2)(1) & = \ln 2 + (2a)(1) \\ 2 & = \ln 2 + 2a \\ 2 - \ln 2 & = 2a \\ {2 - \ln 2 \over 2} & = a \\ \\ \therefore a & = 1 - {1 \over 2}\ln 2 \end{align}

(a)(i)

\begin{align} \text{When } & t = 20, \\ m & = 32e^{-0.02(20)} \\ & = 21.450 \\ & \approx 21.5 \end{align}

(a)(ii)

\begin{align} \text{When } & t = 0, \\ m & = 32e^{-0.02(0)} \\ & = 32(1) \\ & = 32 \end{align}
\begin{align} \text{When } & m = 16, \\ 16 & = 32e^{-0.02t} \\ {16 \over 32} & = e^{-0.02t} \\ 0.5 & = e^{-0.02t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln 0.5 & = \ln e^{-0.02t} \\ \ln 0.5 & = (-0.02t)(\ln e) \phantom{00000000} [\text{Power law}] \\ \ln 0.5 & = (-0.02t) (1) \\ \ln 0.5 & = -0.02t \\ {\ln 0.5 \over -0.02} & = t \\ 34.657 & = t \\ 34.7 & \approx t \end{align}

(b)

\begin{align} \log_3 5 + \log_3 (4 + x) & = \log_3 (10 - x) + 2\log_4 2 \\ [\text{Product law}] \phantom{0000000000} \log_3 [5(4 + x)] & = \log_3 (10 - x) + 2\log_4 2 \\ \log_3 (20 + 5x) & = \log_3 (10 - x) + 2\log_4 2 \\ \log_3 (20 + 5x) - \log_3 (10 - x) & = 2\log_4 2 \\ [\text{Quotient law}] \phantom{00000000}\log_3 \left(20 + 5x \over 10 - x\right) & = 2\log_4 2 \\ \log_3 \left(20 + 5x \over 10 - x\right) & = \log_4 2^2 \phantom{00000000} [\text{Power law}] \\ \log_3 \left(20 + 5x \over 10 - x\right) & = \log_4 4 \\ \log_3 \left(20 + 5x \over 10 - x\right) & = 1 \\ \\ \implies {20 + 5x \over 10 - x} & = 3^1 \phantom{000000000000} [\text{Change to index form}] \\ {20 + 5x \over 10 - x} & = 3 \\ 20 + 5x & = 3(10 - x) \\ 20 + 5x & = 30 - 3x \\ 5x + 3x & = 30 - 20 \\ 8x & = 10 \\ x & = {10 \over 8} \\ & = {5 \over 4} \end{align}

(c)

\begin{align} 10^{2x + 1} + 7(10^x) & = 26 \\ [ a^{m + n} = a^m \times a^n] \phantom{00000000} 10^{2x} \times 10^1 + 7(10^x) & = 26 \\ 10(10^{2x}) + 7(10^x) & = 26 \\ [ a^{mn} = (a^m)^n] \phantom{000000000.} 10(10^x)^2 + 7(10^x) & = 26 \\ \\ \text{Let } & u = 10^x, \\ 10u^2 + 7u & = 26 \\ 10u^2 + 7u - 26 & = 0 \\ (10u - 13)(u + 2) & = 0 \end{align} \begin{align} 10u - 13 & = 0 \phantom{00}&\text{or}\phantom{0000} u + 2 & = 0 \\ 10u & = 13 & u & = - 2 \\ u & = {13 \over 10} \\ \\ \text{Since } & u = 10^x, & \text{Since } & u = 10^x, \\ 10^x & = {13 \over 10} & 10^x & = -2 \text{ (Reject, since } 10^x > 0) \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 10^x & = \lg {13 \over 10} \\ [\text{Power law}] \phantom{0000} x \lg 10 & = \lg {13 \over 10} \\ x (1) & = \lg {13 \over 10} \\ x & = \lg {13 \over 10} \\ \\ \text{Comparing with } & x = \lg a, \\ a & = {13 \over 10} \end{align}

Graph for the entire question (Graph on Desmos):

(i)

(ii)

\begin{align} xe^{2x} & = \sqrt{e^3} \\ xe^{2x} & = (e^3)^{1 \over 2} \\ xe^{2x} & = e^{3 \over 2} \\ x & = {e^{3 \over 2} \over e^{2x}} \\ x & = e^{{3 \over 2} - 2x} \phantom{00000000000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln x & = \ln e^{{3 \over 2} - 2x} \\ & = \left({3 \over 2} - 2x\right) (\ln e) \phantom{000000} [\text{Power law}] \\ & = \left({3 \over 2} - 2x\right) (1) \\ & = {3 \over 2} - 2x \\ \\ \therefore \ln x & = -2x + {3 \over 2} \end{align}

(iii)

(iv)

From the graph, there is only 1 point of intersection between the curve and the line. Thus, the solution in (iii) is the only solution.