(a)

\begin{align} \text{Gradient} & = {5 - 1 \over 0 - 6} \\ m & = -{2 \over 3} \\ \\ \text{Vertical intercept} & = 5 \\ c & = 5 \\ \\ \text{Linear-form: } \phantom{0} Y & = -{2 \over 3}X + 5 \\ \\ \text{Since } X = xy & \text{ and } Y = y, \\ y & = -{2 \over 3}xy + 5 \\ 3y & = -2xy + 15 \\ 3y + 2xy & = 15 \\ y(3 + 2x) & = 15 \\ y & = {15 \over 3 + 2x} \end{align}

(b)

\begin{align} \text{Gradient} & = {7 - 1 \over 2 - (-1)} \\ m & = 2 \\ \\ \text{Let the vertical} & \text{ intercept be } (0, c). \\ \\ \text{Gradient} & = {7 - c \over 2 - 0} \\ 2 & = {7 - c \over 2} \\ 2(2) & = 7 - c \\ 4 & = 7 - c \\ c & = 7 - 4 \\ & = 3 \\ \\ \text{Linear-form: } \phantom{0} Y & = 2X + 3 \\ \\ \\ \text{Since } Y = \lg y & \text{ and } X = \lg x, \\ \lg y & = 2\lg x + 3 \\ \lg y & = \lg x^2 + 3 \phantom{00000} [\text{Power law (logarithms)}] \\ \lg y - \lg x^2 & = 3 \\ \lg \left(y \over x^2\right) & = 3 \phantom{00000000000} [\text{Quotient law (logarithms)}] \\ \log_{10} \left(y \over x^2\right) & = 3 \\ {y \over x^2} & = 10^3 \phantom{000000000} [\text{Change to index form}] \\ {y \over x^2} & = 1000 \\ y & = 1000x^2 \end{align}

\begin{align} \text{Gradient, } m & = 0.75 \\ & = {3 \over 4} \\ \\ \text{Vertical intercept, } c & = -0.5 \\ & = -{1 \over 2} \\ \\ \text{Linear-form: } \phantom{0} Y & = {3 \over 4}X - {1 \over 2} \\ \\ \text{Since } Y = {1 \over y} & \text{ and } X = {1 \over x}, \\ {1 \over y} & = {3 \over 4}\left(1 \over x\right) - {1 \over 2} \\ {1 \over y} & = {3 \over 4x} - {1 \over 2} \\ {1 \over y} - {3 \over 4x} & = - {1 \over 2} \\ {3 \over 4x} - {1 \over y} & = {1 \over 2} \\ 4 \left( {3 \over 4x} - {1 \over y} \right) & = 4\left(1 \over 2\right) \\ {3 \over x} - {4 \over y} & = 2 \\ {3 \over x} + {-4 \over y} & = 2 \\ \\ \therefore a = 3, & \phantom{0} b = -4 \end{align}

\begin{align} y & = \sqrt{ax + b} \\ y^2 & = (\sqrt{ax + b})^2 \\ y^2 & = ax + b \\ \\ \therefore \text{Plot } y^2 & \text{ against } x \\ \\ Y & = y^2 \\ X & = x \\ \text{Gradient, } m & = a \\ \text{Vertical intercept, } c & = b \end{align}

(i)

$ t $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $
$ \ln y $	$ 2.50 $	$ 1.95 $	$ 1.39 $	$ 0.83 $	$ 0.26 $

(ii)

\begin{align} y & = Ae^{-bt} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln (Ae^{-bt}) \\ \ln y & = \ln A + \ln e^{-bt} \phantom{000000} [\text{Product law (logarithms)}] \\ \ln y & = \ln A + (-bt)(\ln e) \phantom{000} [\text{Power law (logarithms)}] \\ \ln y & = \ln A + (-bt)(1) \\ \\ \text{Linear form: } \ln y & = -bt + \ln A \\ \\ \text{where } Y & = \ln y \\ X & = t \\ m & = -b \\ c & = \ln A \\ \\ \text{From }& \text{the graph,} \\ \text{Vertical intercept, } \ln A & = 3.05 \\ \log_e A & = 3.05 \\ A & = e^{3.05} \phantom{000000} [\text{Change to index form}] \\ & = 21.11534 \\ & \approx 21.1 \\ \\ \text{Gradient, } -b & = {2.225 - 1.1 \over 1.5 - 3.5} \\ - b & = -0.5625 \\ b & \approx 0.56 \end{align}

(i)

$ {1 \over v} $	$ 0.05 $	$ 0.059 $	$ 0.067 $	$ 0.077 $	$ 0.08 $
$ {1 \over u} $	$ 0.05 $	$ 0.04 $	$ 0.033 $	$ 0.025 $	$ 0.02 $

(ii)

\begin{align} {1 \over u} + {1 \over v} & = {1 \over f} \\ \\ \text{Since } Y = {1 \over u} & \text{ and } X = {1 \over v}, \\ Y + X & = {1 \over f} \\ \text{Linear form: } \phantom{0} Y & = - X + {1 \over f} \\ \\ \text{where } m & = - 1 \\ c & = {1 \over f} \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } {1 \over f} & = 0.1 \\ 1 & = 0.1f \\ {1 \over 0.1} & = f \\ 10 & = f \end{align}

(i)

\begin{align} x^my^n & = 200 \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg (x^m y^n) & = \lg 200 \\ [\text{Product law (logarithms)}] \phantom{00000} \lg x^m + \lg y^n & = \lg 200 \\ [\text{Power law (logarithms)}] \phantom{0000/} m\lg x + n\lg y & = \lg 200 \\ n \lg y & = -m \lg x + \lg 200 \\ \\ \text{Linear form: } \phantom{0} \lg y & = -{m \over n} \lg x + {\lg 200 \over n} \\ \\ \text{where } Y & = \lg y \\ X & = \lg x \\ \text{Gradient} & = -{m \over n} \\ c & = {\lg 200 \over n} \\ \\ \therefore \text{Plot } \lg y & \text{ against } \lg x \end{align}

$ \lg x $	$ 0 $	$ 0.48 $	$ 0.70 $	$ 0.85 $
$ \lg y $	$ 1.15 $	$ 0.43 $	$ 0.10 $	$ -0.12 $

(ii)

\begin{align} \text{From } & \text{the graph,} \\ \text{Vertical intercept, } {\lg 200 \over n} & = 1.15 \\ \lg 200 & = 1.15 n \\ \\ n & = {\lg 200 \over 1.15} \\ & = 2.0008 \\ & \approx 2 \\ \\ \\ \text{Gradient, } -{m \over n} & = {0.7 - 0.34 \over 0.3 - 0.54} \\ -{m \over n} & = -1.5 \\ -{m \over (2.0008)} & = -1.5 \\ -m & = 2.0008(-1.5) \\ -m & = -3.0012 \\ m & \approx 3 \end{align}

(iii)

\begin{align} \text{When } & y = 10, \\ \lg y & = \lg 10 \\ & = 1 \\ \\ \text{From the graph and } & \text{when } \lg y = 1, \\ \lg x & = 0.1 \\ \log_{10} x & = 0.1 \\ x & = 10^{0.1} \phantom{000000} [\text{Change to index form}] \\ & = 1.2589 \\ & \approx 1.26 \end{align}

(i)

\begin{align} \text{Gradient, } m & = {1 - (-4) \over 10 - 0} \\ & = {1 \over 2} \\ \\ \text{Vertical intercept, } c & = -4 \\ \\ Y & = mX + c \\ Y & = {1 \over 2}X - 4 \\ \\ \text{Since } Y = {1 \over y} & \text{ and } X = x, \\ {1 \over y} & = {1 \over 2}x - 4 \\ {1 \over y} & = {x \over 2} - {8 \over 2} \\ {1 \over y} & = {x - 8 \over 2} \\ 2 & = y(x - 8) \\ {2 \over x - 8} & = y \\ {2 \over x - 8} \times {2 \over 2} & = y \\ {4 \over 2x - 16} & = y \\ \\ \therefore h = 4, & \phantom{0} k = -16 \end{align}

(ii)

\begin{align} \text{On the graph, } & \text{when } x = r, \\ \\ {1 \over y} & = -2 \\ 1 & = - 2y \\ -{1 \over 2} & = y \\ \\ y & = {4 \over 2x - 16} \\ \\ \text{When } x = r & \text{ and } y = -{1 \over 2}, \\ -{1 \over 2} & = {4 \over 2(r) - 16} \\ -(2r - 16) & = 2(4) \\ -2r + 16 & = 8 \\ - 2r & = 8 - 16 \\ - 2r & = -8 \\ r & = {-8 \over -2} \\ r & = 4 \end{align}

(a)

\begin{align} y & = {x \over px + q} \\ y(px + q) & = x \\ px + q & = {x \over y} \\ \\ \therefore {x \over y} & = px + q \\ \\ \text{Linear-form: } \phantom{0} Y & = pX + q \\ \\ \text{where } Y & = {x \over y} \\ X & = x \\ m & = p \\ c & = q \end{align}

(b)

\begin{align} y & = pq^{-x} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln (pq^{-x}) \\ \ln y & = \ln p + \ln q^{-x} \phantom{00000} [\text{Product law (logarithms)}] \\ \ln y & = \ln p + (-x)(\ln q) \phantom{00} [\text{Power law (logarithms)}] \\ \ln y & = (-\ln q)x + \ln p \\ \\ \text{Linear-form: } \phantom{0} Y & = (-\ln q)X + \ln p \\ \\ \text{where } Y & = \ln y \\ X & = x \\ m & = -\ln q \\ c & = \ln p \end{align}

(c)

\begin{align} e^y & = px^2 - qx \\ e^y & = x(px - q) \\ {e^y \over x} & = px - q \\ \\ \text{Linear-form: } \phantom{0} Y & = pX - q \\ \\ \text{where } Y & = {e^y \over x} \\ X & = x \\ m & = p \\ c & = -q \end{align}

(i)

$ x $	$ 1 $	$ 2 $	$ 3 $	$ 4 $
$ (xy - x^2) $	$ 3 $	$ 1 $	$ -0.99 $	$ -3 $

(ii)(a)

\begin{align} \text{From the graph and } & x = 1.5, \\ xy - x^2 & = 2 \\ (1.5)y - (1.5)^2 & = 2 \\ 1.5y - 2.25 & = 2 \\ 1.5y & = 4.25 \\ y & = {4.25 \over 1.5} \\ y & = 2.833333 \\ y & \approx 2.83 \end{align}

(ii)(b)

\begin{align} axy - b & = a(x^2 + bx) \\ {axy \over a} - {b \over a} & = x^2 + bx \\ xy - {b \over a} & = x^2 + bx \\ xy & = x^2 + bx + {b \over a} \\ xy - x^2 & = bx + {b \over a} \\ \\ \text{Since } Y = xy & - x^2 \text{ and } X = x, \\ Y & = bX + {b \over a} \\ \\ \therefore m = b, & \phantom{.} c = {b \over a} \\ \\ \text{From } & \text{the graph,} \\ \\ \text{Gradient, } b & = {4 - (-2) \over 0.5 - 3.5} \\ b & = -2 \\ \\ \text{Vertical intercept, } {b \over a} & = 5 \\ b & = 5a \\ -2 & = 5a \\ {-2 \over 5} & = a \\ -0.4 & = a \end{align}

(i)

$ x $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $	$ 6 $	$ 7 $
$ \lg y $	$ 1.193 $	$ 0.991 $	$ 0.785 $	$ 0.477 $	$ 0.380 $	$ 0.176 $	$ -0.045 $

(ii)(a)

\begin{align} \text{Flawed reading, } y & = 3.0 \\ \\ \text{Corrected reading, } \lg y & = 0.58 \\ \log_{10} y & = 0.58 \\ \\ y & = 10^{0.58} \phantom{000000} [\text{Change to index form}] \\ & = 3.80189 \\ & \approx 3.80 \end{align}

(ii)(b)

\begin{align} y & = Ka^{-x} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg y & = \lg (Ka^{-x}) \\ \lg y & = \lg K + \lg a^{-x} \phantom{00000} [\text{Product law (logarithms)}] \\ \lg y & = \lg K + (-x)\lg a \phantom{000} [\text{Power law (logarithms)}] \\ \lg y & = (- \lg a)x + \lg K \\ \text{Since } Y = \lg y & \text{ and } X = x, \\ \\ m = -\lg a & \text{ and } c = \lg K \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } \lg K & = 1.4 \\ \log_{10} K & = 1.4 \\ K & = 10^{1.4} \phantom{00000000} [\text{Change to index form}] \\ & = 25.1188 \\ & \approx 25.1 \\ \\ \text{Gradient, } -\lg a & = {1.3 - 0.06 \over 0.5 - 6.5} \\ -\lg a & = -0.20666 \\ \lg a & = 0.20666 \\ \log_{10} a & = 0.20666 \\ a & = 10^{0.20666} \\ & = 1.60938 \\ & \approx 1.61 \end{align}

(ii)(c)

\begin{align} \text{When } & y = 10, \\ \lg y & = \lg 10 \\ & = 1 \\ \\ \text{From the graph, } & \text{when } \lg y = 1, \\ x & = 1.9 \\ & \approx 2 \end{align}

(i)

\begin{align} y & = \ln (ax^2 + b) \\ y & = \log_e (ax^2 + b) \\ e^y & = ax^2 + b \\ \\ \text{Linear-form: } \phantom{0} Y & = aX + b \\ \\ \text{where } Y & = e^y \\ X & = x^2 \\ m & = a \\ c & = b \\ \\ \therefore \text{Plot } e^y & \text{ against } x^2 \end{align}

$ x^2 $	$ 0.04 $	$ 0.16 $	$ 0.36 $	$ 0.64 $	$ 1.00 $
$ e^y $	$ 1.38 $	$ 1.62 $	$ 2.02 $	$ 2.58 $	$ 3.30 $

(ii)

\begin{align} \text{From } & \text{the graph,} \\ \\ \text{Vertical-intercept, } b & = 1.3 \\ \\ \text{Gradient, } a & = {2.9 - 1.5 \over 0.8 - 0.1} \\ & = 2 \end{align}

(iii)(a)

\begin{align} \text{When } & y = \ln 3, \\ e^y & = e^{\ln 3} \\ & = 3 \\ \\ \text{From the graph, } & \text{when } e^y = 3, \\ x^2 & = 0.85 \\ x & = \pm 0.92195 \\ & \approx \pm 0.92 \end{align}

(iii)(b)

\begin{align} \text{When } & x = 0.1, \\ x^2 & = (0.1)^2 \\ & = 0.01 \\ \\ \text{From the graph, } & \text{when } x^2 = 0.01, \\ e^y & = 1.325 \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln e^y & = \ln 1.325 \\ [\text{Power law}] \phantom{000000} y (\ln e) & = \ln 1.325 \\ y(1) & = \ln 1.325 \\ y & = \ln 1.325 \\ & = 0.28141 \\ & \approx 0.28 \end{align}

(i)

$ t $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $
$ \lg P $	$ 1.14 $	$ 1.40 $	$ 1.67 $	$ 1.93 $	$ 2.20 $

(ii)

\begin{align} P & = kc^t \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg P & = \lg (kc^t) \\ \lg P & = \lg k + \lg c^t \phantom{00000} [\text{Product law (logarithms)}] \\ \lg P & = \lg k + t(\lg c) \phantom{0000} [\text{Power law (logarithms)}] \\ \lg P & = (\lg c) t + \lg k \\ \\ \text{Linear-form: } \phantom{0} Y & = (\lg c) X + \lg k \\ \\ \text{where } Y & = \lg P \\ X & = t \\ m & = \lg c \\ \text{Vertical intercept} & = \lg k \\ \\ \text{From } & \text{the graph,} \\ \\ \text{Vertical intercept, } \lg k & = 0.875 \\ \log_{10} k & = 0.875 \\ k & = 10^{0.875} \phantom{00000000} [\text{Change to index form}] \\ & = 7.49894 \\ & \approx 7.50 \\ \\ \text{Gradient, } \lg c & = {2.15 - 1 \over 4.8 - 0.5} \\ \lg c & = 0.26744 \\ \log_{10} c & = 0.26744 \\ c & = 10^{0.26744} \\ & = 1.85114 \\ & \approx 1.85 \end{align}

(iii)

\begin{align} \lg P & = (\lg c)t + \lg k \\ \\ \text{Since } \lg c & = 0.26744 \text{ and } \lg k = 0.875, \\ \lg P & = 0.26744t + 0.875 \\ \\ \text{When } & t = 10, \\ \lg P & = 0.26744(10) + 0.875 \\ \lg P & = 3.5494 \\ \log_{10} P & = 3.5494 \\ P & = 10^{3.5494} \\ & = 3543.235 \\ & = 3.543235 \times 10^3 \\ & \approx 3.5 \times 10^3 \end{align}