(a)

\begin{align} \log_2 8 & = \log_2 2^3 \\ & = 3\log_2 2 \phantom{00000000} [\text{Power law}] \\ & = 3(1) \phantom{00000000000} [\log_a a = 1] \\ & = 3 \end{align}

(b)

\begin{align} \log_3 27 + \log_3 {1 \over 9} & = \log_3 3^3 + \log_3 {1 \over 3^2} \\ & = \log_3 3^3 + \log_3 3^{-2} \\ & = 3\log_3 3 + (-2) \log_3 3 \phantom{00000000} [\text{Power law}] \\ & = 3(1) + (-2)(1) \phantom{00000000000000} [\log_a a = 1 ] \\ & = 1 \end{align}

(c)

\begin{align} \log_6 54 - 2\log_6 3 & = \log_6 54 - \log_6 3^2 \phantom{00000000} [\text{Power law}] \\ & = \log_6 54 - \log_6 9 \\ & = \log_6 {54 \over 9} \phantom{000000000000000} [\text{Quotient law}] \\ & = \log_6 6 \\ & = 1 \end{align}

(a)

\begin{align} \log_5 4 + 2\log_5 3 - 3\log_5 2 & = \log_5 4 + \log_5 3^2 - \log_5 2^3 \phantom{00000000} [\text{Power law}] \\ & = \log_5 4 + \log_5 9 - \log_5 8 \\ & = \log_5 (4 \times 9) - \log_5 8 \phantom{0000000000000} [\text{Product law}] \\ & = \log_5 36 - \log_5 8 \\ & = \log_5 {36 \over 8} \phantom{00000000000000000000.00} [\text{Quotient law}] \\ & = \log_5 {9 \over 2} \end{align}

(b) Since lg is equivalent to log₁₀, all 3 terms have the same base

\begin{align} \lg {8 \over 75} - 2\lg {3 \over 5} + 4\lg {3 \over 2} & = \lg {8 \over 75} - \lg \left(3 \over 5\right)^2 + \lg \left(3 \over 2\right)^4 \phantom{00000000} [\text{Power law}] \\ & = \lg {8 \over 75} - \lg {9 \over 25} + \lg {81 \over 16} \\ & = \lg { {8 \over 75} \over {9 \over 25} } + \lg {81 \over 16} \phantom{0000000000000000000.} [\text{Quotient law}] \\ & = \lg {8 \over 27} + \lg {81 \over 16} \\ & = \lg \left( {8 \over 27} \times {81 \over 16} \right) \phantom{0000000000000000000.} [\text{Product law}] \\ & = \lg {3 \over 2} \end{align}

(c)

\begin{align} 1 + 2\log_3 5 - \log_3 {1 \over 27} & = 1 + \log_3 5^2 - \log_3 {1 \over 27} \phantom{00000000} [\text{Power law}] \\ & = 1 + \log_3 25 - \log_3 {1 \over 27} \\ & = 1 + \log_3 {25 \over {1 \over 27}} \phantom{0000000000000000} [\text{Quotient law}] \\ & = 1 + \log_3 675 \\ & = \log_3 3 + \log_3 675 \phantom{000000000000} [\log_a a = 1] \\ & = \log_3 (3 \times 675) \phantom{000000000000000} [\text{Product law}] \\ & = \log_3 2025 \end{align}

(a)

\begin{align} \log_5 7 & = {\log_{10} 7 \over \log_{10} 5} \\ & = {\lg 7 \over \lg 5} \\ & = 1.2090 \\ & \approx 1.21 \end{align}

(b)

\begin{align} \log_{1 \over 2} 5.3 & = {\log_e 5.3 \over \log_e {1 \over 2}} \\ & = {\ln 5.3 \over \ln {1 \over 2}} \\ & = -2.4059 \\ & \approx -2.41 \end{align}

(a)

\begin{align} \log_3 5 \times \log_5 27 & = \log_3 5 \times {\log_3 27 \over \log_3 5} \phantom{00000000} [\text{Change of base}] \\ & = {\log_3 5 \over 1} \times {\log_3 27 \over \log_3 5} \\ & = \log_3 27 \\ & = \log_3 3^3 \\ & = 3\log_3 3 \phantom{0000000000000000} [\text{Power law}] \\ & = 3(1) \phantom{0000000000000000000} [\log_a a = 1] \\ & = 3 \end{align}

(b)

\begin{align} {\log_5 4 \times \log_2 10 \over \log_{25} \sqrt{10}} & = {\log_5 4 \times {\log_5 10 \over \log_5 2} \over \log_{25} \sqrt{10}} \phantom{00000000} [\text{Change of base}] \\ & = {\log_5 2^2 \times {\log_5 10 \over \log_5 2} \over \log_{25} \sqrt{10}} \\ & = {2\log_5 2 \times {\log_5 10 \over \log_5 2} \over \log_{25} \sqrt{10}} \phantom{00000000} [\text{Power law}] \\ & = { {2\log_5 2 \over 1} \times {\log_5 10 \over \log_5 2} \over \log_{25} \sqrt{10}} \\ & = {2\log_5 10 \over \log_{25} \sqrt{10}} \\ & = {2\log_5 10 \over {\log_5 \sqrt{10} \over \log_5 25}} \phantom{00000000000000} [\text{Change of base}] \\ & = {2\log_5 10 \over {\log_5 \sqrt{10} \over \log_5 5^2}} \\ & = {2\log_5 10 \over {\log_5 \sqrt{10} \over 2\log_5 5}} \phantom{00000000000000} [\text{Power law}] \\ & = {2\log_5 10 \over {\log_5 \sqrt{10} \over 2(1)}} \\ & = 2\log_5 10 \div {\log_5 \sqrt{10} \over 2} \\ & = {2\log_5 10 \over 1} \times {2 \over \log_5 \sqrt{10}} \\ & = {4\log_5 10 \over \log_5 \sqrt{10}} \\ & = {4\log_5 10 \over \log_5 10^{1 \over 2}} \\ & = {4\log_5 10 \over {1 \over 2}\log_5 10} \phantom{00000000000000} [\text{Power law}] \\ & = {4 \over {1 \over 2}} \\ & = 8 \end{align}

(a)

\begin{align} y & = 100x^{1.5} \\ \\ \text{Taking } & \lg \text{ of both sides,} \\ \lg y & = \lg 100x^{1.5} \\ \lg y & = \lg 100 + \lg x^{1.5} \phantom{00000000} [\text{Product law}] \\ \lg y & = \lg 10^2 + \lg x^{1.5} \\ \lg y & = 2\lg 10 + 1.5 \lg x \phantom{0000000} [\text{Power law}] \\ \lg y & = 2(1) + 1.5 \lg x \phantom{000000000} [\lg 10 = \log_{10} 10 = 1] \\ \lg y & = 2 + 1.5\lg x \\ \lg y & = 1.5\lg x + 2 \\ \\ \text{Compar} & \text{ing with } \lg y = mX + c, \\ m & = 1.5, X = \lg x \text{ and } c = 2 \end{align}

(b)

\begin{align} y & = 0.1 (1000)^x \\ y & = {1 \over 10}(1000)^x \\ y & = (10^{-1})(1000)^x \\ y & = (10^{-1})(10^3)^x \\ y & = (10^{-1})(10^{3x}) \phantom{0000000} [ (a^m)^n = a^{mn}] \\ y & = 10^{-1 + 3x} \phantom{00000000000} [ (a^m)(a^n) = a^{m + n}] \\ \\ \text{Taking } & \lg \text{ of both sides,} \\ \lg y & = \lg 10^{-1 + 3x} \\ \lg y & = (-1 + 3x) \lg 10 \phantom{00000} [\text{Power law}] \\ \lg y & = (-1 + 3x)(1) \phantom{0000000} [\lg 10 = \log_{10} = 10 = 1] \\ \lg y & = -1 + 3x \\ \lg y & = 3x - 1 \\ \\ \text{Compar} & \text{ing with } \lg y = mX + c, \\ m & = 3, X = x \text{ and } c = -1 \end{align}

(a)

\begin{align} \log_a 8 - 2\log_a 4 & = \log_a 8 - \log_a 4^2 \phantom{000000} [\text{Power law}] \\ & = \log_a 8 - \log_a 16 \\ & = \log_a {8 \over 16} \phantom{000000000000} [\text{Quotient law}] \\ & = \log_a {1 \over 2} \end{align}

(b)

\begin{align} 2\log_a 5 - 3 \log_a 2 + \log_a 4 & = \log_a 5^2 - \log_a 2^3 + \log_a 4 \phantom{000000} [\text{Power law}] \\ & = \log_a 25 - \log_a 8 + \log_a 4 \\ & = \log_a {25 \over 8} + \log_a 4 \phantom{0000000000000} [\text{Quotient law}] \\ & = \log_a \left({25 \over 8} \times 4\right) \phantom{000000000000.00} [\text{Product law}] \\ & = \log_a {25 \over 2} \end{align}

(c)

\begin{align} 3\log_a 2 - 4 + 2\log_a a^3 & = 3\log_a 2 - 4(1) + 2\log_a a^3 \\ & = 3\log_a 2 - 4\log_a a + 2\log_a a^3 \phantom{000000} [\log_a a = 1] \\ & = \log_a 2^3 - \log_a a^4 + \log_a (a^3)^2 \phantom{000000.} [\text{Power law}] \\ & = \log_a 8 - \log_a a^4 + \log_a a^6 \\ & = \log_a {8 \over a^4} + \log_a a^6 \phantom{000000000000000.} [\text{Quotient law}] \\ & = \log_a \left({8 \over a^4} \times a^6\right) \phantom{00000000000000000} [\text{Product law}] \\ & = \log_a \left( {8 \over a^4} \times {a^6 \over 1} \right) \\ & = \log_a (8a^2) \end{align}

(i)

\begin{align} {\log_a 25 \over \log_a 3a} & = {\log_a 5^2 \over \log_a 3a} \\ & = {2\log_a 5 \over \log_a 3a} \phantom{0000000000} [\text{Power law}] \\ & = {2\log_a 5 \over \log_a 3 + \log_a a} \phantom{00000} [\text{Product law}] \\ & = {2\log_a 5 \over \log_a 3 + 1} \phantom{000000000} [\log_a a = 1] \\ & = {2(0.699) \over (0.477) + 1} \phantom{00000000} [\text{Use values provided}] \\ & = 0.94651 \\ & \approx 0.947 \end{align}

(ii)

\begin{align} \log_a (5a^2) & = \log_a 5 + \log_a a^2 \phantom{00000000} [\text{Product law}] \\ & = \log_a 5 + 2\log_a a \phantom{0000000.} [\text{Power law}] \\ & = \log_a 5 + 2(1) \phantom{0000000000} [\log_a a = 1] \\ & = \log_a 5 + 2 \\ & = 0.699 + 2 \phantom{000000000000.} [\text{Use value provided}] \\ & = 2.699 \end{align}

(iii)

\begin{align} \log_a {9 \over 5a} & = \log_a 9 - \log_a 5a \phantom{000000000000} [\text{Quotient law}] \\ & = \log_a 3^2 - \log_a 5a \\ & = 2\log_a 3 - \log_a 5a \phantom{00000000000} [\text{Power law}] \\ & = 2\log_a 3 - (\log_a 5 + \log_a a) \phantom{000.} [\text{Product law}] \\ & = 2\log_a 3 - \log_a 5 - \log_a a \\ & = 2\log_a 3 - \log_a 5 - 1 \phantom{00000000.} [\log_a a = 1] \\ & = 2(0.477) - (0.699 + 1) \phantom{0000000} [\text{Use values provided}] \\ & = -0.745 \end{align}

(i)

\begin{align} \log_4 45 & = \log_4 (3 \times 15) \\ & = \log_4 3 + \log_4 15 \phantom{000000} [\text{Product law} ] \\ & = \log_4 3 + \log_4 (3 \times 5) \\ & = \log_4 3 + (\log_4 3 + \log_4 5) \\ & = a + a + b \\ & = 2a + b \end{align}

(ii)

\begin{align} \log_4 20 & = \log_4 (4 \times 5) \\ & = \log_4 4 + \log_4 5 \phantom{00000000} [\text{Product law}] \\ & = 1 + \log_4 5 \phantom{000000000000} [\log_a a = 1] \\ & = 1 + b \end{align}

(iii)

\begin{align} \log_{75} 16 & = {\log_4 16 \over \log_4 75} \phantom{00000000000000} [\text{Change of base}] \\ & = {\log_4 16 \over \log_4 (3 \times 25)} \\ & = {\log_4 16 \over \log_4 3 + \log_4 25} \phantom{0000000} [\text{Product law}] \\ & = {\log_4 4^2 \over \log_4 3 + \log_4 5^2} \\ & = {2\log_4 4 \over \log_4 3 + 2\log_4 5} \phantom{0000000} [\text{Power law}] \\ & = {2(1) \over \log_4 3 + 2\log_4 5} \phantom{0000000} [\log_a a = 1] \\ & = {2 \over \log_4 3 + 2\log_4 5} \\ & = {2 \over a + 2b} \end{align}

(iv)

\begin{align} \log_4 3 & = a & \log_4 5 & = b \\ \\ \implies 3 & = 4^a & 5 & = 4^b \end{align}
\begin{align} 375 & = 3 \times 125 \\ & = 3 \times 5 \times 25 \\ & = 3 \times 5 \times 5 \times 5 \\ & = 4^a \times 4^b \times 4^b \times 4^b \\ & = 4^{a + b} \times 4^b \times 4^b \phantom{000000} [ a^m \times a^n = a^{m + n} ] \\ & = 4^{a + b + b} \times 4^b \\ & = 4^{a + 2b} \times 4^b \\ & = 4^{a + 2b + b} \\ & = 4^{a + 3b} \end{align}

Due to complexity, simplify each term individually first: \begin{align} 2\lg 3\sqrt{x} & = \lg (3\sqrt{x})^2 \phantom{000000} [\text{Power law}] \\ & = \lg 9x \\ & = \lg 9 + \lg x \phantom{00000} [\text{Product law}] \\ \\ \lg {4 \over 3x^2} & = \lg \left( {4 \over 3} \times {1 \over x^2} \right) \\ & = \lg {4 \over 3} + \lg {1 \over x^2} \phantom{000000000} [\text{Product law}] \\ & = \lg {4 \over 3} + \lg x^{-2} \\ & = \lg {4 \over 3} + (-2)\lg x \phantom{000000.} [\text{Power law}] \\ & = \lg {4 \over 3} - 2 \lg x \\ \\ \log_{100} 4x^3 & = {\log_{10} 4x^3 \over \log_{10} 100} \phantom{00000000} [\text{Change of base}] \\ & = {\lg 4x^3 \over \lg 100} \\ & = {\lg 4x^3 \over \lg 10^2} \\ & = {\lg 4x^3 \over 2\lg 10} \\ & = {\lg 4x^3 \over 2(1)} \\ & = {\lg 4x^3 \over 2} \\ & = {1 \over 2} \lg 4x^3 \\ & = {1 \over 2}(\lg 4 + \lg x^3) \phantom{00000000} [\text{Product law}] \\ & = {1 \over 2}(\lg 2^2 + \lg x^3) \\ & = {1 \over 2}(2\lg 2 + 3\lg x) \phantom{000000} [\text{Power law}] \\ & = \lg 2 + {3 \over 2}\lg x \\ \\ \\ \therefore 2 \lg 3\sqrt{x} - \lg {4 \over 3x^2} + \log_{100} 4x^3 & = (\lg 9 + \lg x) - (\lg {4 \over 3} - 2\lg x) + (\lg 2 + {3 \over 2}\lg x) \\ & = \lg 9 + \lg x - \lg {4 \over 3} + 2\lg x + \lg 2 + {3 \over 2}\lg x \\ & = \lg x + 2\lg x + {3 \over 2}\lg x + \lg 9 - \lg {4 \over 3} + \lg 2 \\ & = \lg x (1 + 2 + {3 \over 2}) + \lg 9 - \lg {4 \over 3} + \lg 2 \\ & = {9 \over 2}\lg x + \lg 9 - \lg {4 \over 3} + \lg 2 \\ & = {9 \over 2}\lg x + \lg {9 \over {4 \over 3}} + \lg 2 \phantom{000000000000000000} [\text{Quotient law}] \\ & = {9 \over 2}\lg x + \lg {27 \over 4} + \lg 2 \\ & = {9 \over 2}\lg x + \lg \left( {27 \over 4} \times 2 \right) \phantom{00000000000000000} [\text{Power law}] \\ & = {9 \over 2}\lg x + \lg {27 \over 2} \end{align}

(i)

\begin{align} \lg \left(100\sqrt{x} \over y^2\right) & = \lg (100\sqrt{x}) - \lg y^2 \phantom{000000000000} [\text{Quotient law}] \\ & = (\lg 100 + \lg \sqrt{x}) - \lg y^2 \phantom{0000000.} [\text{Product law}] \\ & = \lg 10^2 + \lg x^{1 \over 2} - \lg y^2 \\ & = 2\lg 10 + {1 \over 2}\lg x - 2\lg y \phantom{0000000.} [\text{Power law}] \\ & = 2(1) +{1 \over 2}\lg x - 2\lg y \phantom{000000000.} [\lg 10 = \log_{10} 10 = 1] \\ & = 2 + {1 \over 2}\lg x - 2\lg y \\ & = 2 + {1 \over 2}p - 2q \end{align}

(ii)

\begin{align} \lg x & = p \\ \log_{10} x & = p \\ \\ \implies x & = 10^p \\ \\ \\ \lg y^x & = x\lg y \phantom{0000000000} [\text{Power law}] \\ & = (10^p) (q) \\ & = q(10^p) \end{align}

\begin{align} \ln K & = \ln a - \ln b + \ln c - {t \over bc} \\ \ln K & = \ln {a \over b} + \ln c - {t \over bc} \phantom{000000000000} [\text{Quotient law}] \\ \ln K & = \ln \left({a \over b} \times c\right) - {t \over bc} \phantom{000000000000} [\text{Product law}] \\ \ln K & = \ln {ac \over b} - {t \over bc} \\ \ln K - \ln {ac \over b} & = -{t \over bc} \\ [\text{Quotient law}] \phantom{00000000000} \ln {K \over {ac \over b}} & = -{t \over bc} \\ \ln {Kb \over ac} & = -{t \over bc} \\ \log_e {Kb \over ac} & = -{t \over bc} \phantom{0000000000000000} [\text{Change to index form}] \\ \\ \implies {Kb \over ac} & = e^{-{t \over bc}} \\ Kb & = (ac)e^{-{t \over bc}} \\ K & = {ac \over b} e^{-{t \over bc}} \phantom{00} \text{ (Shown)} \end{align}

(a)

\begin{align} \lg (y + 1) & = 2 - {1 \over 2} \lg x \\ \lg (y + 1) & = 2 - \lg x^{1 \over 2} \phantom{00000000} [\text{Power law}] \\ \lg (y + 1) & = 2 - \lg \sqrt{x} \\ \lg (y +1) + \lg \sqrt{x} & = 2 \\ [\text{Product law}] \phantom{0000000000} \lg [\sqrt{x}(y + 1)] & = 2 \\ \log_{10} [\sqrt{x}(y + 1)] & = 2 \\ \\ \implies \sqrt{x}(y + 1) & = 10^2 \phantom{0000000000000} [\text{Change to index form}] \\ \sqrt{x}(y + 1) & = 100 \\ y + 1 & = {100 \over \sqrt{x}} \\ \therefore y & = {100 \over \sqrt{x}} - 1 \end{align}

(b)

\begin{align} 2\log_3 y - 4 & = 3 \log_3 (x + 2) \\ \log_3 y^2 - 4 & = \log_3 (x + 2)^3 \phantom{00000000} [\text{Power law}] \\ \log_3 y^2 - \log_3 (x + 2)^3 & = 4 \\ [\text{Quotient law}] \phantom{0000000000} \log_3 {y^2 \over (x + 2)^3} & = 4 \\ \\ \implies {y^2 \over (x + 2)^3} & = 3^4 \phantom{000000000000} [\text{Change to index form}] \\ {y^2 \over (x + 2)^3} & = 81 \\ y^2 & = 81(x + 2)^3 \\ \\ y & = \pm\sqrt{81 (x + 2)^3} \\ & = \pm\sqrt{81}\sqrt{(x + 2)^3} \\ & = \pm 9 \sqrt{(x + 2)^3} \\ & = \pm 9 (x + 2)^{3 \over 2} \\ & = 9(x + 2)^{3 \over 2} \phantom{0}\text{ or }\phantom{0} -9(x + 2)^{3 \over 2} \phantom{0} \left( \text{Reject, since } \log_3 \left[ -9(x + 2)^{3 \over 2} \right] \text{ is undefined} \right) \\ \\ \therefore y & = 9(x + 2)^{3 \over 2} \end{align}

(c)

\begin{align} 3 + \log_2 (x + y) & = \log_2 (x - 2y) \\ \log_2 (x + y) - \log_2 (x - 2y) & = - 3 \\ [\text{Quotient law}] \phantom{0000000000} \log_2 {x + y \over x - 2y} & = - 3 \\ \\ \implies {x + y \over x - 2y} & = 2^{-3} \phantom{0000000000} [\text{Change to index form}] \\ {x + y \over x - 2y} & = {1 \over 8} \\ 8(x + y) & = x - 2y \\ 8x + 8y & = x - 2y \\ 10y & = - 7x \\ y & = -{7 \over 10} x \end{align}

\begin{align} \log_2 (y + 1) & = 2\log_2 x + c \\ \log_2 (y + 1) & = \log_2 x^2 + c \phantom{00000000} [\text{Power law}] \\ \log_2 (y + 1) - \log_2 x^2 & = c \\ [\text{Quotient law}] \phantom{0000000000} \log_2 {y + 1 \over x^2} & = c \\ \\ \implies {y + 1 \over x^2} & = 2^c \phantom{0} \text{ --- (1)} \\ \\ \\ \text{When } x = 2 & \text{ and } y = 3, \\ {3 + 1 \over 2^2} & = 2^c \\ 1 & = 2^c \\ 2^0 & = 2^c \\ \\ \therefore c & = 0 \\ \\ \text{Substitute } & c = 0 \text{ into (1),} \\ {y + 1 \over x^2} & = 2^0 \\ {y + 1 \over x^2} & = 1 \\ y + 1 & = x^2 \\ \therefore y & = x^2 - 1 \end{align}

Since m = lg 2, we have to change the term to base 10: \begin{align} \log_8 5 & = {\log_{10} 5 \over \log_{10} 8} \phantom{00000000} [\text{Change of base}] \\ & = {\lg 5 \over \lg 8} \\ & = {\lg 5 \over \lg 2^3} \\ & = {\lg 5 \over 3\lg 2} \phantom{000000000} [\text{Power law}] \\ & = {\lg {10 \over 2} \over 3\lg 2} \\ & = {\lg 10 - \lg 2 \over 3\lg 2} \phantom{0000} [\text{Quotient law}] \\ & = {1 - \lg 2 \over 3\lg 2} \phantom{0000000} [\lg 10 = \log_{10} 10 = 1] \\ & = {1 - m \over 3m} \end{align}

(i)

Rationalise the denominator: \begin{align} \text{R.H.S} & = {1 \over \sqrt{3} + \sqrt{2}} \times {\sqrt{3} - \sqrt{2} \over \sqrt{3} - \sqrt{2}} \\ & = {\sqrt{3} - \sqrt{2} \over (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\ & = {\sqrt{3} - \sqrt{2} \over (\sqrt{3})^2 - (\sqrt{2})^2} \\ & = {\sqrt{3} - \sqrt{2} \over 3 - 2} \\ & = {\sqrt{3} - \sqrt{2} \over 1} \\ & = \sqrt{3} - \sqrt{2} \\ & = \text{L.H.S} \end{align}

(ii)

\begin{align} \lg (\sqrt{3} - \sqrt{2}) & = \lg \left(1 \over \sqrt{3} + \sqrt{2}\right) \phantom{00000000} [\text{Apply result from part (i)}] \\ & = \lg (\sqrt{3} + \sqrt{2})^{-1} \\ & = (-1) \lg (\sqrt{3} + \sqrt{2}) \phantom{00000000} [\text{Power law}] \\ & = -\lg (\sqrt{3} + \sqrt{2}) \end{align}

(iii)

Using part (i) as inspiration, $\sqrt{9} - \sqrt{5}$ can be expressed as: \begin{align} \sqrt{9} - \sqrt{5} & = (\sqrt{9} - \sqrt{5}) \times {\sqrt{9} + \sqrt{5} \over \sqrt{9} + \sqrt{5}} \\ & = {(\sqrt{9} - \sqrt{5})(\sqrt{9} + \sqrt{5}) \over \sqrt{9} + \sqrt{5}} \\ & = {(\sqrt{9})^2 - (\sqrt{5})^2 \over \sqrt{9} + \sqrt{5}} \\ & = {9 - 5 \over \sqrt{9} + \sqrt{5}} \\ & = {4 \over \sqrt{9} + \sqrt{5}} \\ \\ \\ \therefore \log_{\sqrt{2}} (\sqrt{9} - \sqrt{5}) & = \log_{\sqrt{2}} \left(4 \over \sqrt{9} + \sqrt{5}\right) \\ & = \log_{\sqrt{2}} 4 - \log_{\sqrt{2}} (\sqrt{9} + \sqrt{5}) \phantom{00000000} [\text{Quotient law}] \\ & = {\log_2 4 \over \log_2 \sqrt{2}} - {\log_2 (\sqrt{9} + \sqrt{5}) \over \log_2 \sqrt{2}} \phantom{0000000} [\text{Change of base}] \\ & = {\log_2 2^2 \over \log_2 2^{1 \over 2}} - {\log_2 (\sqrt{9} + \sqrt{5}) \over \log_2 2^{1 \over 2}} \\ & = {2\log_2 2 \over {1 \over 2} \log_2 2 } - {\log_2 (\sqrt{9} + \sqrt{5}) \over {1 \over 2}\log_2 2} \\ & = {2(1) \over {1 \over 2}(1)} - {\log_2 (\sqrt{9} + \sqrt{5}) \over {1 \over 2}(1)} \\ & = {2 \over {1 \over 2}} - {\log_2 (\sqrt{9} + \sqrt{5}) \over {1 \over 2} } \\ & = 4 - 2\log_2 (\sqrt{9} + \sqrt{5}) \\ & = 4 - 2k \\ & = 2(2 - k) \end{align}

(i)

\begin{align} p & = a^{\log_a x} \\ \\ \text{Take } \log_a & \text{ of both sides,} \\ \log_a p & = \log_a a^{\log_a x} \\ \log_a p & = (\log_a x)(\log_a a) \phantom{00000000} [\text{Power law}] \\ \log_a p & = (\log_a x)(1) \\ \log_a p & = \log_a x \\ \\ \therefore p & = x \text{ (Shown)} \end{align}

(ii)

\begin{align} \log_4 3 & = q \\ \\ 3 & = 4^q \\ \\ \therefore 3^{1 \over q} & = (3)^{1 \over q} \\ & = (4^q)^{1 \over q} \\ & = 4^{q \times {1 \over q}} \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = 4^1 \\ & = 4 \end{align}

\begin{align} (\lg 5)^2 + \lg 2 \lg 50 & = (\lg 5)^2 + \lg 2 \lg (5 \times 10) \\ & = (\lg 5)^2 + \lg 2 (\lg 5 + \lg 10) \phantom{00000000} [\text{Product law}] \\ & = (\lg 5)^2 + \lg 2 (\lg 5 + 1) \\ & = (\lg 5)^2 + \lg 2 \lg 5 + \lg 2 \\ & = \lg 5 (\lg 5 + \lg 2) + \lg 2 \phantom{00000000000} [\text{Factorise } \lg 5 ] \\ & = \lg 5 [\lg (5 \times 2)] + \lg 2 \phantom{000000000000} [\text{Product law}] \\ & = \lg 5 (\lg 10) + \lg 2 \\ & = \lg 5 (1) + \lg 2 \\ & = \lg 5 + \lg 2 \\ & = \lg (5 \times 2) \\ & = \lg 10 \\ & = 1 \end{align}

(i)

\begin{align} \lg {1 \over 2} & = \lg 1 - \lg 2 \\ \\ \lg {2 \over 3} & = \lg 2 - \lg 3 \\ \\ \lg {3 \over 4} & = \lg 3 - \lg 4 \\ \\ \lg {998 \over 999} & = \lg 998 - \lg 999 \\ \\ \lg {999 \over 1000} & = \lg 999 - \lg 1000 \end{align}
\begin{align} \require{cancel} &\therefore \lg {1 \over 2} + \lg {2 \over 3} + \lg {3 \over 4} + ... + \lg {998 \over 999} + \lg {999 \over 1000} \\ & = (\lg 1 - \lg 2) + (\lg 2 - \lg 3) + (\lg 3 - \lg 4) + ... + (\lg 998 - \lg 999) + (\lg 999 - \lg 1000) \\ & = \lg 1 \cancel{-\lg 2} \cancel{+\lg 2} \cancel{-\lg 3} \cancel{+\lg 3} \cancel{-\lg 4} + ... + \cancel{\lg 998} \cancel{-\lg 999} \cancel{+\lg 999} - \lg 1000 \\ & = \lg 1 - \lg 1000 \\ & = 0 - 3 \\ & = -3 \end{align}

(ii)(a)

\begin{align} u_n & = \left(1 - {1 \over n} \right)^2 \\ u_n & = \left( {n \over n} - {1 \over n} \right)^2 \\ u_n & = \left( n - 1 \over n \right)^2 \\ \\ \text{Taking } \lg & \text{ on both sides,} \\ \lg u_n & = \lg \left( n - 1 \over n \right)^2 \\ \lg u_n & = 2 \lg \left( n - 1 \over n \right) \phantom{0000000000} [\text{Power law}] \\ \\ \text{When } & n = 4, \\ \lg u_4 & = 2 \lg \left(4 - 1 \over 4 \right) \\ & = 2 \lg {3 \over 4} \text{ (Shown)} \\ \\ \text{When } & n = 10^k, \\ \lg u_{(10^k)} & = 2 \lg \left( 10^k - 1 \over 10^k \right) \end{align}

(ii)(b)

\begin{align} \lg u_n & = 2 \lg \left( n - 1 \over n \right) \\ & = 2 [\lg (n - 1) - \lg n] \\ & = 2\lg (n - 1) - 2 \lg n \\ \\ \lg u_2 & = 2\lg (2 - 1) - 2 \lg 2 \\ & = 2\lg 1 - 2\lg 2 \\ \\ \lg u_3 & = 2\lg (3 - 1) - 2 \lg 3 \\ & = 2\lg 2 - 2\lg 3 \\ \\ \lg u_4 & = 2\lg (4 - 1) - 2 \lg 4 \\ & = 2\lg 3 - 2 \lg 4 \\ \\ \lg u_{(10^k)} & = 2 \lg (10^k - 1) - 2 \lg 10^k \end{align}

Using the values above, \begin{align} \require{cancel} & \phantom{=/} \lg u_2 + \lg u_3 + \lg u_4 + ...+ \lg u_{(10^k)} \\ & = (2\lg 1 - 2\lg 2) + (2\lg 2 - 2\lg 3) + (2\lg 3 - 2\lg 4) + ... + [2\lg(10^k - 1) - 2\lg 10^k] \\ & = 2\lg 1 \cancel{- 2\lg 2} \cancel{+ 2\lg 2} \cancel{- 2\lg 3} \cancel{+ 2\lg 3} \cancel{-2\lg 4} + ... + \cancel{2\lg (10^k - 1)} - 2\lg 10^k \\ & = 2\lg 1 - 2\lg 10^k \\ & = 2(0) - 2\lg 10^k \\ & = 0 - 2\lg 10^k \\ & = -2\lg 10^k \\ & = -2k \lg 10 \\ & = -2k (1) \\ & = -2k \end{align}

\begin{align} \lg {1 \over 2} & = -0.30102 \\ \\ \\ 3 & > 2 \\ 3 \lg {1 \over 2} & < 2 \lg {1 \over 2} \phantom{000000} [\text{Corrected step}] \\ \lg \left(1 \over 2\right)^3 & < \lg \left(1 \over 2\right)^2 \\ \left(1 \over 2\right)^3 & < \left(1 \over 2\right)^2 \\ {1 \over 8} & < {1 \over 4} \end{align}

First part

\begin{align} \lg [(-1)(-2)] & = \lg 2 \\ \\ \lg (-1) + \lg (-2) & \text{ is undefined} \\ \\ \\ \therefore \lg [(-1)(-2)] & \ne \lg (-1) + \lg (-2) \end{align}

(i)

\begin{align} x & = -1, y = 2 \end{align}

(ii)

\begin{align} x & = -1, r = 5 \end{align}