(a)

\begin{align} \text{Let } & y = 0, \\ 0 & = \log_2 x \\ \\ \implies 2^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(b)

\begin{align} \text{Let } & y = 0, \\ 0 & = \log_6 x \\ \\ \implies 6^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(c)

\begin{align} \text{Let } & y = 0, \\ 0 & = \log_{0.2} x \\ \\ \implies (0.2)^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(d)

\begin{align} \text{Let } & y = 0, \\ 0 & = \log_{1 \over 4} x \\ \\ \implies \left(1 \over 4\right)^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(e)

\begin{align} y & = \lg x \\ y & = \log_{10} x \\ \\ \text{Let } & y = 0, \\ 0 & = \log_{10} x \\ \\ \implies (10)^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(f)

\begin{align} y & = \ln x \\ y & = \log_{e} x \\ \\ \text{Let } & y = 0, \\ 0 & = \log_{e} x \\ \\ \implies e^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

(i)

\begin{align} \text{When } & \text{H}^+ = 5.6 \times 10^{-7}, \\ \text{pH} & = -\log_{10} (5.6 \times 10^{-7}) \\ & = -\lg (5.6 \times 10^{-7}) \\ & = 6.2518 \\ \\ \therefore \text{Colour of} & \text{ flower is pink} \end{align}

(ii)

\begin{align} \text{When } & \text{H}^+ = 7.8 \times 10^{-6}, \\ \text{pH} & = -\log_{10} (7.8 \times 10^{-6}) \\ & = -\lg (7.8 \times 10^{-6}) \\ & = 5.1079 \\ \\ \therefore \text{Colour of} & \text{ flower is blue} \end{align}

(i) Initial temperature refers to the temperature of the liquid at the start (i.e. x = 0)

\begin{align} \text{When } & x = 0, \\ T & = 85(0.96)^0 \\ & = 85(1) \\ & = 85^\circ \text{C} \end{align}

(ii)

\begin{align} \text{When } & x = 15, \\ T & = 85(0.96)^{15} \\ & = 46.077 \\ & \approx 46.1 \end{align}

(iii)

\begin{align} \text{When } & T = 30, \\ 30 & = 85(0.96)^x \\ {30 \over 85} & = (0.96)^x \\ {6 \over 17} & = (0.96)^x \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg {6 \over 17} & = \lg (0.96)^x \\ \lg {6 \over 17} & = x \lg 0.96 \phantom{0000000000} [\text{Power law} ] \\ { \lg {6 \over 17} \over \lg 0.96} & = x \\ 25.512 & = x \\ \\ \therefore x & \approx 25.5 \end{align}

(i)

\begin{align} \text{When } & t = 0, \\ N & = 100(1.65)^0 \\ & = 100(1) \\ & = 100 \end{align}

(ii)

\begin{align} \text{When } & t = 4, \\ N & = 100(1.65)^4 \\ & = 741.20 \\ & \approx 741 \end{align}

(iii)

\begin{align} \text{When } & N = 400, \\ 400 & = 100(1.65)^t \\ {400 \over 100} & = (1.65)^t \\ 4 & = (1.65)^t \\ \\ \text{Taking } \lg & \text{ of both sides,} \\ \lg 4 & = \lg (1.65)^t \\ \lg 4 & = t\lg 1.65 \phantom{0000000000} [\text{Power law}] \\ {\lg 4 \over \lg 1.65} & = t \\ \\ t & = {\lg 4 \over \lg 1.65} \\ & = 2.7682 \\ & \approx 3 \end{align}

(i)

\begin{align} \text{When } & t = 10000, \\ R & = 100e^{-0.0004279(10000)} \\ & = 1.3856 \\ & \approx 1.39 \text{ g} \end{align}

(ii) Since the beginning amount of radium is 100 g, half the amount of radium is 50 g

\begin{align} \text{When } & R = 50, \\ 50 & = 100e^{-0.0004279t} \\ {50 \over 100} & = e^{-0.0004279t} \\ 0.5 & = e^{-0.0004279t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln 0.5 & = \ln e^{-0.0004279t} \\ \ln 0.5 & = -0.0004279t(\ln e) \phantom{0000000000} [\text{Power law}] \\ \ln 0.5 & = -0.0004279t(1) \\ \ln 0.5 & = -0.0004279t \\ {\ln 0.5 \over -0.0004279} & = t \\ \\ t & = {\ln 0.5 \over -0.0004279} \\ & = 1619.8 \\ & \approx 1620 \text{ years} \end{align}

(i)

\begin{align} \text{When } & T = 1.5, \\ I & = 0.87^{1.5} \\ & = 0.81148 \\ & \approx 0.811 \end{align}

(ii)

\begin{align} \text{When } & I = 0.5, \\ 0.5 & = 0.87^T \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 0.5 & = \lg 0.87^T \\ \lg 0.5 & = T \lg 0.87 \phantom{0000000000} [\text{Power law}] \\ {\lg 0.5 \over \lg 0.87} & = T \\ \\ T & = {\lg 0.5 \over \lg 0.87} \\ & = 4.9772 \\ & = 4.98 \text{ mm} \end{align}

(i)

\begin{align} y & = \lg x \\ \\ \text{Let } & y = 0, \\ 0 & = \lg x \\ 0 & = \log_{10} x \\ \\ \implies 10^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \\ \\ \\ y & = \ln x \\ \\ \text{Let } & y = 0, \\ 0 & = \ln x \\ 0 & = \log_{e} x \\ \\ \implies e^0 & = x \phantom{00000000} [\text{Change to index form}] \\ 1 & = x \\ \\ x\text{-int} & \text{ercept is } 1 \end{align}

To differentiate between the two curves, find the $y$-coordinate when $x = 4$
\begin{align} \text{When } & x = 4, \\ y & = \lg 4 \\ & \approx 0.602 \\ \\ \\ \text{When } & x = 4, \\ y & = \ln 4 \\ & \approx 1.39 \end{align}

(ii)

(a) From the graph, $ \lg x > 0 $ when $ x > 1 $

(b) From the graph, $ \lg x \le 0 $ when $ 0 < x \le 1 $

(c) From the graph, $ \ln x > \lg x $ when $ x > 1 $

(i)

\begin{align} \text{When } & t = 5, \\ A & = 5000(1.04)^5 \\ & = 6083.2 \\ & \approx \$6083 \end{align}

(ii)

\begin{align} \text{When }& A = 8000, \\ 8000 & = 5000(1.04)^t \\ {8000 \over 5000} & = (1.04)^t \\ 1.6 & = (1.04)^t \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg 1.6 & = \lg (1.04)^t \\ \lg 1.6 & = t \lg 1.04 \phantom{0000000000} [\text{Power law}] \\ {\lg 1.6 \over \lg 1.04} & = t \\ \\ t & = {\lg 1.6 \over \lg 1.04} \\ & = 11.983 \\ & \approx 12 \end{align}

(iii)

\begin{align} A & = 5000(1.04)^t \\ {A \over 5000} & = (1.04)^t \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg \left({A \over 5000}\right) & = \lg (1.04)^t \\ \lg \left({A \over 5000}\right) & = t \lg 1.04 \phantom{00000000000000.} [\text{Power law}] \\ \\ t & = {\lg \left({A \over 5000}\right) \over \lg 1.04} \\ t & = {\log_{10} \left({A \over 5000}\right) \over \log_{10} 1.04} \\ t & = \log_{1.04} \left({A \over 5000}\right) \phantom{00000000} [\text{Change of base}] \end{align}

(i)

\begin{align} \text{When } & t = 4, \\ S & = 75 - 6\ln [(4) + 1] \\ & = 75 - 6\ln 5 \\ & = 65.343 \\ & = 65.3 \end{align}

(ii)

\begin{align} \text{When } & S = 62, \\ 62 & = 75 - 6\ln (t + 1) \\ 62 - 75 & = -6 \ln (t + 1) \\ -13 & = -6 \ln (t + 1) \\ {-13 \over -6} & = \ln (t + 1) \\ {13 \over 6} & = \ln (t + 1) \\ {13 \over 6} & = \log_e (t + 1) \\ \\ \implies e^{13 \over 6} & = t + 1 \phantom{0000000000} [\text{Change to index form}] \\ \\ t & = e^{13 \over 6} - 1 \\ & = 7.7291 \\ & \approx 8 \end{align}

(iii)

\begin{align} S & = 75 - 6\ln (t + 1) \\ S - 75 & = -6 \ln (t + 1) \\ 75 - S & = 6 \ln (t + 1) \\ {1 \over 6}(75 - S) & = \ln (t + 1) \\ {1 \over 6}(75 - S) & = \log_e (t + 1) \\ & \downarrow \\ e^{{1 \over 6}(75 - S)} & = t + 1 \phantom{0000000000} [\text{Change to index form}] \\ \\ t & = e^{{1 \over 6}(75 - S)} - 1 \end{align}

(i)

\begin{align} \text{When } & t = 6, \\ y & = {1500 \over 1 + 1499e^{-0.85(6)}} \\ & = 147.94 \\ & \approx 148 \end{align}

(ii) 40% of 1500 is 600. So the school would cancel class if 600 or more students are infected

\begin{align} \text{When } & y = 600, \\ 600 & = {1500 \over 1 + 1499e^{-0.85t}} \\ 600(1 + 1499e^{-0.85t}) & = 1500 \\ 1 + 1499e^{-0.85t} & = {1500 \over 600} \\ 1 + 1499e^{-0.85t} & = 2.5 \\ 1499e^{-0.85t} & = 1.5 \\ e^{-0.85t} & = 1.5 \div 1499 \\ e^{-0.85t} & = {3 \over 2998} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln e^{-0.85t} & = \ln \left(3 \over 2998\right) \\ [\text{Power law}] \phantom{00000000} -0.85t (\ln e) & = \ln \left(3 \over 2998\right) \\ -0.85t (1) & = \ln \left(3 \over 2998\right) \\ -0.85t & = \ln \left(3 \over 2998\right) \\ t & = \ln \left(3 \over 2998\right) \div -0.85 \\ & = 8.1259 \\ & \approx 9 \end{align}

\begin{align} t & = -2.35 \ln \left(T - 24.5 \over 12.4 \right) \\ \\ \text{When } & T = 28.8, \\ t & = -2.35 \ln \left(28.8 - 24.5 \over 12.4\right) \\ & = 2.4888 \text{ hours} \\ & = 2 \text{ hours } (0.4888 \times 60) \text{ mins} \\ & \approx 2 \text{ hours } 29 \text{ mins} \\ \\ \text{Estimated time} & = 7:00 \text{ am } - \text{2 hours 29 mins} \\ & = 4 : 31 \text{ am} \\ \\ \\ \text{When } & T = 28.0, \\ t & = -2.35 \ln \left(28.0 - 24.5 \over 12.4\right) \\ & = 2.9725 \text{ hours} \\ & = 2 \text{ hours } (0.9725 \times 60) \text{ mins} \\ & \approx 2 \text{ hours } 58 \text{ mins} \\ \\ \text{Estimated time} & = 7:29 \text{ am } - \text{2 hours 58 mins} \\ & = 4 : 31 \text{ am} \\ \\ \\ \therefore \text{Estimated time } & \text{of death is } 4:31 \text{ am} \end{align}

(i)

\begin{align} 2 \log_4 x & = 1 \\ \log_4 x & = {1 \over 2} \\ x & = 4^{1 \over 2} \\ x & = 2 \end{align}

(ii)

(iii)

$$ 0 < x \le 2 $$

(i)

\begin{align} T & = T_0 + Ce^{-kt} \\ \\ 29 \text{ mins} & = {29 \over 60} \text{ hours} \\ \\ 28.0 & = 24.5 + Ce^{-k \left(t_1 + {29 \over 60} \right)} \end{align}

(ii)

\begin{align} 28.8 & = 24.5 + Ce^{-k t_1} \\ 4.3 & = Ce^{-k t_1} \phantom{00} \text{--- (1)} \\ \\ 28.0 & = 24.5 + Ce^{-k \left(t_1 + {29 \over 60} \right)} \\ 3.5 & = Ce^{-k \left(t_1 + {29 \over 60} \right)} \phantom{00} \text{--- (2)} \\ \\ (2) & \div (1), \\ {3.5 \over 4.3} & = { Ce^{-k \left(t_1 + {29 \over 60} \over Ce^{-k t_1} \right)}} \\ {35 \over 43} & = e^{- k \left(t_1 + {29 \over 60} \right) + k t_1} \\ {35 \over 43} & = e^{-{29 \over 60}k} \phantom{00} \text{ (Shown)} \\ \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {35 \over 43} & = \ln e^{-{29 \over 60}k} \\ \ln {35 \over 43} & = -{29 \over 60}k \\ {29 \over 60}k & = - \ln {35 \over 43} \\ k & = - \ln {35 \over 43} \div {29 \over 60} \\ k & = 0.4259 \\ \\ \\ T & = 24.5 + Ce^{-0.4259 t} \\ \\ \text{When } & t = 0 \text{ and } T = 36.9, \\ 36.9 & = 24.5 + Ce^{-0.4259(0)} \\ 36.9 & = 24.5 + C(1) \\ 12.4 & = C \\ \\ \\ \therefore k & \approx 0.426, C = 12.4 \end{align}

(iii)

\begin{align} T & = 24.5 + 12.4e^{-0.4259 t} \\ \\ T - 24.5 & = 12.4 e^{-0.4259 t} \\ {T - 24.5 \over 12.4} & = e^{-0.4259 t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln \left( {T - 24.5 \over 12.4} \right) & = \ln e^{-0.4259t} \\ \ln \left( {T - 24.5 \over 12.4} \right) & = -0.4259 t \\ \\ t & = {1 \over - 0.4259} \ln \left( {T - 24.5 \over 12.4} \right) \\ t & \approx -2.35 \ln \left( {T - 24.5 \over 12.4} \right) \end{align}

(i)

\begin{align} & y = 2 \ln x \text{ is only defined for } x > 0 \\ \\ & y = \ln x^2 \text{ is defined for all real values of } x \text{ except } x = 0 \\ \\ & y = (\ln x)^2 \text{ is only defined for } x > 0 \text{ and there will be no negative values of } y, \text{ i.e. } y \ge 0 \end{align}

(ii)