(a)

\begin{align} \log_5 (x + 1) & = \log_5 3 \\ \\ \therefore x + 1 & = 3 \\ x & = 2 \end{align}

(b)

\begin{align} \log_2 (x - 1) & = \log_2 (4x - 7) \\ \\ \therefore x - 1 & = 4x - 7 \\ -3x & = - 6 \\ x & = 2 \end{align}

(c)

\begin{align} \log_3 x + \log_3 (x + 2) & = 1 \\ [\text{Product law}] \phantom{00000000} \log_3 [(x)(x + 2)] & = 1 \\ \log_3 (x^2 + 2x) & = 1 \\ \\ \implies x^2 + 2x & = 3^1 \phantom{00000000} [\text{Change to index form}] \\ x^2 + 2x & = 3 \\ x^2 + 2x - 3 & = 0 \\ (x - 1)(x + 3) & = 0 \\ \\ x - 1 = 0 \phantom{00}&\text{or} \phantom{00} x + 3 = 0 \\ x = 1 \phantom{00}&\phantom{or00+3} x = - 3 \text{ (Reject, since } \log_3 (-3) \text{ and } \log_3 (-1) \text{ are undefined}) \end{align}

(a)

\begin{align} \log_x 25 + \log_x 5 & = 3 \\ [\text{Product law}] \phantom{00000000} \log_x (25 \times 5) & = 3 \\ \log_x 125 & = 3 \\ \\ \implies 125 & = x^3 \phantom{00000000} [\text{Change to index form}] \\ \sqrt[3]{125} & = x \\ 5 & = x \end{align}

(b)

\begin{align} 3 \log_x 2 + \log_x 18 & = 2 \\ [\text{Power law}] \phantom{00000000} \log_x 2^3 + \log_x 18 & = 2 \\ \log_x 8 + \log_x 18 & = 2 \\ [\text{Product law}] \phantom{00000000000} \log_x (8 \times 18) & = 2 \\ \log_x 144 & = 2 \\ \\ \implies 144 & = x^2 \phantom{00000000} [\text{Change to index form}] \\ \pm \sqrt{144} & = x \\ \pm 12 & = x \\ \\ \therefore x = 12 \phantom{0} & \text{ or } -12 \text{ (Reject, since } \log_{-12} 2 \text{ and } \log_{-12} 18 \text{ are undefined}) \end{align}

(a)

\begin{align} \lg [(x + 2)(x - 2)] & = \lg (2x - 1) \\ \\ \therefore (x + 2)(x - 2) & = 2x - 1 \\ x^2 - 4 & = 2x - 1 \\ x^2 - 2x - 3 & = 0 \\ (x - 3)(x + 1) & = 0 \\ \\ x - 3 = 0 \phantom{00} & \text{or} \phantom{00} x + 1 = 0 \\ x = 3 \phantom{00} & \phantom{or00+1} x = - 1 \text{ (Reject, since } \lg (-3) \text{ is undefined}] \end{align}

(b)

\begin{align} \log_2 [(x - 2)(8 - x)] - \log_2 (x - 5) & = 3 \\ [\text{Quotient law}] \phantom{00000000} \log_2 \left[ (x - 2)(8 - x) \over x - 5 \right] & = 3 \\ \\ \implies { (x - 2)(8 - x) \over x - 5} & = 2^3 \phantom{00000000} [\text{Change to index form}] \\ {(x - 2)(8 - x) \over x - 5} & = 8 \\ (x - 2)(8 - x) & = 8(x - 5) \\ 8x - x^2 - 16 + 2x & = 8x - 40 \\ - x^2 + 2x + 24 & = 0 \\ x^2 - 2x - 24 & = 0 \\ (x - 6)(x + 4) & = 0 \\ \\ x - 6 = 0 \phantom{00}& \text{or} \phantom{00} x + 4 = 0 \\ x = 6 \phantom{00}&\phantom{or00+4} x = -4 \text{ (Reject, since } \log_2 (-72) \text{ is undefined}) \end{align}

(c)

\begin{align} \log_2 (x - 1)^2 & = 2 + \log_2 (x + 2) \\ \log_2 (x - 1)^2 - \log_2 (x + 2) & = 2 \\ [\text{Quotient law}] \phantom{00000000} \log_2 \left[{(x - 1)^2 \over x + 2}\right] & = 2 \\ \\ \implies {(x - 1)^2 \over x + 2} & = 2^2 \\ {(x - 1)^2 \over x + 2} & = 4 \\ (x - 1)^2 & = 4(x + 2) \\ x^2 - 2x + 1 & = 4x + 8 \\ x^2 - 6x - 7 & = 0 \\ (x - 7)(x + 1) & = 0 \\ \\ x - 7 = 0 \phantom{00}&\text{or}\phantom{00} x + 1 = 0 \\ x = 7 \phantom{00}&\phantom{or00+1} x = - 1 \end{align}

(a)

\begin{align} \log_3 x & = 9 \log_x 3 \\ \log_3 x & = 9 \left( \log_3 3 \over \log_3 x \right) \phantom{00000000} [\text{Change of base}] \\ \log_3 x & = \left({9 \over 1}\right) \left( 1 \over \log_3 x \right) \\ \log_3 x & = {9 \over \log_3 x} \\ \\ \text{Let } & u = \log_3 x, \\ u & = {9 \over u} \\ u^2 & = 9 \\ u & = \pm \sqrt{9} \\ & = \pm 3 \end{align} \begin{align} \log_3 x & = 3 \phantom{00}&\text{or}\phantom{00} \log_3 x & = - 3 \\ \\ \implies x & = 3^3 & x & = 3^{-3} \\ & = 27 & & = {1 \over 27} \end{align}

(b)

\begin{align} \log_3 x + 2 & = 3\log_x 3 \\ \log_3 x + 2 & = 3 \left( \log_3 3 \over \log_3 x \right) \phantom{00000000} [\text{Change of base}] \\ \log_3 x + 2 & = 3 \left( 1 \over \log_3 x \right) \\ \log_3 x + 2 & = {3 \over \log_3 x} \\ \\ \text{Let } & u = \log_3 x, \\ u + 2 & = {3 \over u} \\ u^2 + 2u & = 3 \\ u^2 + 2u - 3 & = 0 \\ (u + 3)(u - 1) & = 0 \end{align} \begin{align} u + 3 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 1 & = 0 \\ u & = -3 & u & = 1 \\ \\ \text{Since } & u = \log_3 x, \\ \log_3 x & = - 3 & \log_3 x & = 1 \\ \\ \implies x & = 3^{-3} & x & = 3^1 \\ & = {1 \over 27} & & = 3 \end{align}

Simplify each equation before solving by substitution: \begin{align} \ln (3x - y) & = \ln 36 - \ln 9 \\ \ln (3x - y) & = \ln {36 \over 9} \phantom{00000000} [\text{Quotient law}] \\ \ln (3x - y) & = \ln 4 \\ \\ \therefore 3x - y & = 4 \\ - y & = 4 - 3x \\ y & = -4 + 3x \phantom{0} \text{ --- (1)} \\ \\ {(e^x)^2 \over e^y} & = e \\ (e^x)^2 & = e(e^y) \\ [ (a^m)^n = a^{mn}] \phantom{00000000} e^{2x} & = e(e^y) \\ e^{2x} & = e^1 (e^y) \\ e^{2x} & = e^{1 + y} \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ \\ \therefore 2x & = 1 + y \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2x & = (-4 + 3x) + 1 \\ 2x & = - 4 + 3x + 1 \\ 2x - 3x & = -4 + 1 \\ - x & = -3 \\ x & = 3 \\ \\ \text{Substitute } & x = 3 \text{ into (1),} \\ y & = -4 + 3(3) \\ & = 5 \end{align}

Simplify each equation before solving by substitution: \begin{align} 2^{p + 2} & = 16(4^{{3 \over 2}q}) \\ 2^{p + 2} & = 16[ (2^2)^{{3 \over 2}q} ] \\ 2^{p + 2} & = 16( 2^{2 \times {3 \over 2}q} ) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ 2^{p + 2} & = (16)(2^{3q}) \\ 2^{p + 2} & = (2^4)(2^{3q}) \\ 2^{p + 2} & = 2^{4 + 3q} \phantom{000000000000} [ (a^m)(a^n) = a^{m + n}] \\ \\ \therefore p + 2 & = 4 + 3q \\ p & = 2 + 3q \phantom{0} \text{ --- (1)} \\ \\ \\ \log_2 6 - \log_2 (15q - 3p) & = 1 \\ [\text{Quotient law}] \phantom{00000000} \log_2 {6 \over 15q - 3p} & = 1 \\ \\ \implies {6 \over 15q - 3p} & = 2^1 \phantom{000000000000000} [\text{Change to index form}] \\ {6 \over 15q - 3p} & = 2 \\ 6 & = 2(15q - 3p) \\ 6 & = 30q - 6p \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 6 & = 30q - 6(2 + 3q) \\ 6 & = 30q - 12 - 18q \\ 18 & = 12q \\ {18 \over 12} & = q \\ {3 \over 2} & = q \\ \\ \text{Substitute } & q = {3 \over 2} \text{ into (1),} \\ p & = 2 + 3\left(3 \over 2\right) \\ & = 2 + {9 \over 2} \\ & = {13 \over 2} \end{align}

(a)

\begin{align} 2\log_5 x + \log_{25} x & = 5 \\ [\text{Change of base}] \phantom{00000000} 2\log_5 x + {\log_5 x \over \log_5 25} & = 5 \\ 2\log_5 x + {\log_5 x \over \log_5 5^2} & = 5 \\ [\text{Power law}] \phantom{00000000} 2\log_5 x + {\log_5 x \over 2\log_5 5} & = 5 \\ 2\log_5 x + {\log_5 x \over 2(1)} & = 5 \\ 2\log_5 x + {\log_5 x \over 2} & = 5 \\ 4\log_5 x + \log_5 x & = 10 \\ 5\log_5 x & = 10 \\ \log_5 x & = {10 \over 5} \\ \log_5 x & = 2 \\ \\ \implies x & = 5^2 \phantom{00000000} [\text{Change to index form}] \\ & = 25 \end{align}

(b)

\begin{align} \log_2 x - \log_4 (x + 6) & = 0 \\ \log_2 x & = \log_4 (x + 6) \\ \log_2 x & = {\log_2 (x + 6) \over \log_2 4} \phantom{00000000} [\text{Change of base}] \\ \log_2 x & = {\log_2 (x + 6) \over \log_2 2^2} \\ \log_2 x & = {\log_2 (x + 6) \over 2\log_2 2} \\ \log_2 x & = {\log_2 (x + 6) \over 2(1)} \\ \log_2 x & = {\log_2 (x + 6) \over 2} \\ 2\log_2 x & = \log_2 (x + 6) \\ [\text{Power law}] \phantom{00000000} \log_2 x^2 & = \log_2 (x + 6) \\ \\ \therefore x^2 & = x + 6 \\ x^2 - x - 6 & = 0 \\ (x - 3)(x + 2) & = 0 \\ \\ x - 3 = 0 \phantom{00}&\text{or}\phantom{000} x + 2 = 0 \\ x = 3 \phantom{00}&\phantom{or000+2} x = - 2 \text{ (Reject, since } \log_2 (-2) \text{ is undefined}) \end{align}

(c)

\begin{align} \log_5 (5 - 4x) & = \log_{\sqrt{5}} (2 - x) \\ \log_5 (5 - 4x) & = {\log_5 (2 - x) \over \log_5 \sqrt{5}} \phantom{00000000} [\text{Change of base}] \\ \log_5 (5 - 4x) & = {\log_5 (2 - x) \over \log_5 5^{1 \over 2}} \\ \log_5 (5 - 4x) & = {\log_5 (2 - x) \over {1 \over 2}\log_5 5} \\ \log_5 (5 - 4x) & = {\log_5 (2 - x) \over {1 \over 2}(1)} \\ \log_5 (5 - 4x) & = {\log_5 (2 - x) \over {1 \over 2}} \\ \log_5 (5 - 4x) & = 2\log_5 (2 - x) \\ \log_5 (5 - 4x) & = \log_5 (2 - x)^2 \phantom{00000000} [\text{Power law}] \\ \\ \therefore 5 - 4x & = (2 - x)^2 \\ 5 - 4x & = 4 - 4x + x^2 \\ 0 & = x^2 - 4x + 4x + 4 - 5 \\ 0 & = x^2 - 1 \\ 0 & = (x - 1)(x + 1) \\ \\ x - 1= 0 \phantom{00}&\text{or} \phantom{00} x + 1 = 0 \\ x = 1 \phantom{00}& \phantom{or00+1} x = - 1 \end{align}

(a)

\begin{align} \log_4 y & = \log_2 x + \log_2 9 - \log_2 3 \\ \log_4 y & = \log_2 (x \times 9) - \log_2 3 \phantom{0000000000} [\text{Product law}] \\ \log_4 y & = \log_2 9x - \log_2 3 \\ \log_4 y & = \log_2 {9x \over 3} \phantom{0000000000000000000.} [\text{Quotient law}] \\ \log_4 y & = \log_2 3x \\ [\text{Change of base}] \phantom{0000000000} {\log_2 y \over \log_2 4} & = \log_2 3x \\ {\log_2 y \over \log_2 2^2} & = \log_2 3x \\ [\text{Power law}] \phantom{00000000.}{\log_2 y \over 2\log_2 2} & = \log_2 3x \\ {\log_2 y \over 2(1)} & = \log_2 3x \\ {\log_2 y \over 2} & = \log_2 3x \\ \log_2 y & = 2\log_2 3x \\ \log_2 y & = \log_2 (3x)^2 \\ \\ \therefore y & = (3x)^2 \\ y & = 9x^2 \end{align}

(b)

\begin{align} \log_9 y + \log_3 y & = 2\log_3 x + 3\log_3 2 \\ \log_9 y + \log_3 y & = \log_3 x^2 + \log_3 2^3 \phantom{00000000} [\text{Power law}] \\ \log_9 y + \log_3 y & = \log_3 x^2 + \log_3 8 \\ \log_9 y + \log_3 y & = \log_3 (x^2 \times 8) \phantom{000000000000} [\text{Product law}] \\ \log_9 y + \log_3 y & = \log_3 8x^2 \\ \log_9 y & = \log_3 8x^2 - \log_3 y \\ \log_9 y & = \log_3 {8x^2 \over y} \phantom{0000000000000.0} [\text{Quotient law}] \\ [\text{Change of base}] \phantom{00000000} {\log_3 y \over \log_3 9} & = \log_3 {8x^2 \over y} \\ {\log_3 y \over \log_3 3^2} & = \log_3 {8x^2 \over y} \\ [\text{Power law}] \phantom{0000000} {\log_3 y \over 2\log_3 3} & = \log_3 {8x^2 \over y} \\ {\log_3 y \over 2(1)} & = \log_3 {8x^2 \over y} \\ {\log_3 y \over 2} & = \log_3 {8x^2 \over y} \\ \log_3 y & = 2\log_3 {8x^2 \over y} \\ \log_3 y & = \log_3 \left(8x^2 \over y\right)^2 \\ \\ \therefore y & = \left(8x^2 \over y\right)^2 \\ y & = {64x^4 \over y^2} \\ y^2(y) & = 64x^4 \\ y^3 & = 64x^4 \\ y & = \sqrt[3]{64x^4} \\ y & = \sqrt[3]{64} \sqrt[3]{x^4} \\ y & = 4 (x^4)^{1 \over 3} \\ y & = 4x^{4 \over 3} \phantom{0000000000} [ (a^m)^n = a^{mn}] \end{align}

(a)

\begin{align} \log_4 (6 - x) - \log_2 8 & = \log_9 3 \\ \log_4 (6 - x) - \log_2 2^3 & = \log_9 3 \\ [\text{Power law}] \phantom{00000000} \log_4 (6 - x) - 3\log_2 2 & = \log_9 3 \\ \log_4 (6 - x) - 3(1) & = \log_9 3 \\ \log_4 (6 - x) - 3 & = \log_9 3 \\ \log_4 (6 - x) - 3 & = {\log_3 3 \over \log_3 9} \phantom{00000000} [\text{Change of base}] \\ \log_4 (6 - x) - 3 & = {1 \over \log_3 3^2} \\ \log_4 (6 - x) - 3 & = {1 \over 2\log_3 3} \\ \log_4 (6 - x) - 3 & = {1 \over 2(1)} \\ \log_4 (6 - x) - 3 & = {1 \over 2} \\ \log_4 (6 - x) & = {1 \over 2} + 3 \\ \log_4 (6 - x) & = {7 \over 2} \\ \\ \implies 6 - x & = 4^{7 \over 2} \phantom{0000000000} [\text{Change to index form}] \\ 6 - x & = 128 \\ -x & = 128 - 6 \\ -x & = 122 \\ x & = -122 \end{align}

(b)

\begin{align} \log_7 (9x + 38) - \log_7 (x + 2) & = \log_9 81 \\ \log_7 (9x + 38) - \log_7 (x + 2) & = \log_9 9^2 \\ \log_7 (9x + 38) - \log_7 (x + 2) & = 2\log_9 9 \phantom{00000000} [\text{Power law}] \\ \log_7 (9x + 38) - \log_7 (x + 2) & = 2(1) \\ \log_7 (9x + 38) - \log_7 (x + 2) & = 2 \\ [\text{Quotient law}] \phantom{00000000} \log_7 \left({9x + 38 \over x + 2}\right) & = 2 \\ \\ \implies {9x + 38 \over x + 2} & = 7^2 \phantom{000000000000.} [\text{Change to index form}] \\ {9x + 38 \over x + 2} & = 49 \\ 9x + 38 & = 49(x + 2) \\ 9x + 38 & = 49x + 98 \\ 9x - 49x & = 98 - 38 \\ -40x & = 60 \\ x & = {60 \over -40} \\ x & = -{3 \over 2} \end{align}

(c)

\begin{align} \log_9 (3x + 1) & = \log_3 x + \log_3 2 \\ \log_9 (3x + 1) & = \log_3 (x \times 2) \phantom{00000000} [\text{Product law}] \\ \log_9 (3x + 1) & = \log_3 2x \\ [\text{Change of base}] \phantom{00000000} {\log_3 (3x + 1) \over \log_3 9} & = \log_3 2x \\ {\log_3 (3x + 1) \over \log_3 3^2} & = \log_3 2x \\ [\text{Power law}] \phantom{00000000} {\log_3 (3x + 1) \over 2\log_3 3} & = \log_3 2x \\ {\log_3 (3x + 1) \over 2(1)} & = \log_3 2x \\ {\log_3 (3x + 1) \over 2} & = \log_3 2x \\ \log_3 (3x + 1) & = 2\log_3 2x \\ \log_3 (3x + 1) & = \log_3 (2x)^2 \\ \\ \therefore 3x + 1 & = (2x)^2 \\ 3x + 1 & = 4x^2 \\ 0 & = 4x^2 - 3x - 1 \\ 0 & = (x - 1)(4x + 1) \\ \\ x - 1 = 0 \phantom{00}&\text{or} \phantom{00} 4x + 1 = 0 \\ x = 1 \phantom{00}& \phantom{or00+1} 4x = - 1 \\ & \phantom{or00+14} x = -{1 \over 4} \phantom{0} \left( \text{Reject, since } \log_3 \left(-{1 \over 4}\right) \text{ is undefined} \right) \end{align}

(d)

\begin{align} 3\log_4 x - \log_{16} x & = 3.75 \\ [\text{Change of base}] \phantom{00000000} 3\log_4 x - {\log_4 x \over \log_4 16} & = 3.75 \\ 3\log_4 x - {\log_4 x \over \log_4 4^2} & = 3.75 \\ [\text{Power law}] \phantom{00000000} 3\log_4 x - {\log_4 x \over 2\log_4 4} & = 3.75 \\ 3\log_4 x - {\log_4 x \over 2(1)} & = 3.75 \\ 3\log_4 x - {\log_4 x \over 2} & = 3.75 \\ 3\log_4 x - {1 \over 2} \log_4 x & = 3.75 \\ \log_4 x \left( 3 - {1 \over 2} \right) & = 3.75 \\ \log_4 x \left(5 \over 2\right) & = 3.75 \\ \log_4 x & = 3.75 \div {5 \over 2} \\ \log_4 x & = 1.5 \\ \\ \implies x & = 4^{1.5} \phantom{00000000} [\text{Change to index form}] \\ x & = 8 \end{align}

(e)

\begin{align} \log_5 x - \log_{25} (x + 10) & = 0.5 \\ [\text{Change of base}] \phantom{00000000} \log_5 x - {\log_5 (x + 10) \over \log_5 25} & = 0.5 \\ \log_5 x - {\log_5 (x + 10) \over \log_5 5^2} & = 0.5 \\ [\text{Power law}] \phantom{00000000} \log_5 x - {\log_5 (x + 10) \over 2\log_5 5} & = 0.5 \\ \log_5 x - {\log_5 (x + 10) \over 2(1)} & = 0.5 \\ \log_5 x - {\log_5 (x + 10) \over 2} & = 0.5 \\ 2\log_5 x - \log_5 (x + 10) & = 1 \phantom{0000000000} [\text{Multiply by 2}] \\ \log_5 x^2 - \log_5 (x + 10) & = 1 \\ [\text{Quotient law}] \phantom{000000000000} \log_5 \left({x^2 \over x + 10}\right) & = 1 \\ \\ \implies {x^2 \over x + 10} & = 5^1 \phantom{000000000} [\text{Change to index form}] \\ {x^2 \over x + 10} & = 5 \\ x^2 & = 5(x + 10) \\ x^2 & = 5x + 50 \\ x^2 - 5x - 50 & = 0 \\ (x - 10)(x + 5) & = 0 \\ \\ x - 10 = 0 \phantom{000}&\text{or}\phantom{00} x + 5 = 0 \\ x = 10 \phantom{00}& \phantom{or00+5} x = - 5 \text{ (Reject, since } \log_5 (-5) \text{ is undefined}) \end{align}

\begin{align} 2 \log_a x & = 1 + \log_a (7x - 10a) \\ \log_a x^2 & = 1 + \log_a (7x - 10a) \phantom{000000} [\text{Power law}] \\ \log_a x^2 - \log_a (7x - 10a) & = 1 \\ \log_a \left(x^2 \over 7x - 10a\right) & = 1 \phantom{00000000000000000000.} [\text{Quotient law}] \\ \\ {x^2 \over 7x - 10a} & = a^1 \\ {x^2 \over 7x - 10a} & = a \\ x^2 & = a(7x - 10a) \\ x^2 & = 7ax - 10a^2 \\ x^2 - 7ax & = -10a^2 \\ \left(x - {7 \over 2}a\right)^2 - \left({7 \over 2}a\right)^2 & = -10a^2 \phantom{000000000000} [\text{Complete the square}] \\ \left(x - {7 \over 2}a\right)^2 - {49 \over 4}a^2 & = -10a^2 \\ \left(x - {7 \over 2}a\right)^2 & = {9 \over 4}a^2 \\ x - {7 \over 2}a & = \pm \sqrt{ {9 \over 4} a^2 } \\ x - {7 \over 2}a & = \pm {3 \over 2}a \\ x & = 5a \text{ or } 2 a \end{align}

\begin{align} 27 \times 3^{\lg x} & = 9^{1 + \lg (x - 20)} \\ 3^3 \times 3^{\lg x} & = 3^{2[1 + \lg (x - 20)]} \\ 3^{3 + \lg x} & = 3^{2 + 2 \lg (x - 20)} \\ \\ 3 + \lg x & = 2 + 2 \lg (x - 20) \\ 1 & = 2 \lg (x - 20) - \lg x \\ 1 & = \lg (x - 20)^2 - \lg x \phantom{000000} [\text{Power law}] \\ 1 & = \lg \left[ (x - 20)^2 \over x \right] \phantom{00000000.} [\text{Quotient law}] \\ 1 & = \log_{10} \left[ (x - 20)^2 \over x \right] \\ \\ 10^1 & = {(x - 20)^2 \over x} \\ 10x & = (x -20)^2 \\ 10x & = x^2 - 2(x)(20) + (20)^2 \\ 10x & = x^2 - 40x + 400 \\ 0 & = x^2 - 50x + 400 \\ 0 & = (x - 40)(x - 10) \end{align} \begin{align} x - 40 & = 0 && \text{ or } & x - 10 & = 0 \\ x & = 40 &&& x & = 10 \text{ (Reject, since } 9^{1 + \lg(x - 20)} \text{ is undefined}) \end{align}

(i)

\begin{align} \log_2 (2y - 3x) - \log_2 3 & = 4 \log_4 2 \\ \log_2 \left( 2y - 3x \over 3 \right) & = 4 \log_4 2 \phantom{0000000000} [\text{Quotient law}] \\ \log_2 \left( 2y - 3x \over 3 \right) & = 4 \left(\log_2 2 \over \log_2 4\right) \phantom{000000} [\text{Change-of-base}] \\ \log_2 \left( 2y - 3x \over 3 \right) & = 4 \left(1 \over \log_2 2^2\right) \\ \log_2 \left( 2y - 3x \over 3 \right) & = 4 \left(1 \over 2\log_2 2\right) \phantom{00000} [\text{Power law}] \\ \log_2 \left( 2y - 3x \over 3 \right) & = 4 \left[1 \over 2(1)\right] \\ \log_2 \left( 2y - 3x \over 3 \right) & = 2 \\ \\ {2y - 3x \over 3} & = 2^2 \\ {2y - 3x \over 3} & = 4 \\ 2y - 3x & = 3(4) \\ 2y - 3x & = 12 \\ 2y & = 3x + 12 \\ y & = {3 \over 2}x + 6 \phantom{00} \text{--- (1)} \\ \\ \\ \log_3 6 + \log_3 (x + y) & = \log_3 (-x) + \log_3 (1 - x) \\ \log_3 [6(x + y)] & = \log_3 [-x(1 - x)] \phantom{000000000} [\text{Product law}] \\ \\ 6(x + y) & = -x(1 - x) \\ 6x + 6y & = -x + x^2 \\ 6y & = -7x + x^2 \phantom{00} \text{--- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 6 \left({3 \over 2}x + 6\right) & = -7x + x^2 \\ 9x + 36 & = -7x + x^2 \\ 0 & = x^2 - 16x - 36 \\ 0 & = (x + 2)(x - 18) \end{align} \begin{align} x + 2 & = 0 && \text{ or } & x - 18 & = 0 \\ x & = -2 &&& x & = 18 \text{ (Reject, since } \log_3 (-x) \text{ is undefined}) \\ \\ \text{Substitute } & \text{into (1),} \\ y & = {3 \over 2}(-2) + 6 \\ y & = 3 \\ \\ \therefore x & = -2, y = 3 \end{align}

(ii) The steps are similar (i) - the difference is that the second solution of x do not need be rejected

\begin{align} x + 2 & = 0 && \text{ or } & x - 18 & = 0 \\ x & = -2 &&& x & = 18 \\ \\ \text{Substitute } & \text{into (1),} &&& \text{Substitute } & \text{into (1),} \\ y & = {3 \over 2}(-2) + 6 &&& y & = {3 \over 2}(18) + 6 \\ y & = 3 &&& y & = 33 \\ \\ \therefore x & = -2, y = 3 &&& \therefore x & = 18, y = 33 \end{align}

(i)

\begin{align} \log_2 (x^3 + 1) - 2 \log_2 x & = \log_2 (x^2 - x + 1) - 2 \\ \log_2 (x^3 + 1) - \log_2 x^2 & = \log_2 (x^2 - x + 1) - 2 \phantom{0000000} [\text{Power rule}] \\ \log_2 (x^3 + 1) - \log_2 x^2 & = \log_2 (x^2 - x + 1) - 2 \log_2 2 \\ \log_2 (x^3 + 1) - \log_2 x^2 & = \log_2 (x^2 - x + 1) - \log_2 2^2 \\ \log_2 \left(x^3 + 1 \over x^2\right) & = \log_2 \left(x^2 - x + 1 \over 2^2\right) \phantom{00000000} [\text{Quotient rule}] \\ \\ {x^3 + 1 \over x^2} & = {x^2 - x + 1 \over 4} \\ 4(x^3 + 1) & = x^2(x^2 - x + 1) \\ 0 & = x^2 (x^2 - x + 1) - 4(x^3 + 1) \\ 0 & = x^2 (x^2 - x + 1) - 4(x + 1)(x^2 - x + 1) \phantom{00000} [a^3 + b^3 = (a + b)(a^2 - ab + b^2)] \\ 0 & = (x^2 - x + 1)[x^2 - 4(x + 1)] \\ 0 & = (x^2 - x + 1)(x^2 - 4x - 4) \end{align} \begin{align} x^2 - x + 1 & = 0 && \text{ or } & x^2 - 4x - 4 & = 0 \\ b^2 - 4ac & = (-1)^2 - 4(1)(1) &&& x & = {-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)} \over 2(1)} \\ & = -3 &&& & = { 4 \pm \sqrt{32} \over 2} \\ \implies & \text{No real roots} &&& & = {4 \pm 4 \sqrt{2} \over 2} \\ & &&& & = 2 + 2 \sqrt{2} \text{ or } 2 - 2\sqrt{2} \text{ (Reject, since } \log_2 x \text{ is undefined}) \end{align}

(ii) The steps are similar to (i) - the difference is that the second solution of x do not need be rejected

\begin{align} x^2 - x + 1 & = 0 && \text{ or } & x^2 - 4x - 4 & = 0 \\ b^2 - 4ac & = (-1)^2 - 4(1)(1) &&& x & = {-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)} \over 2(1)} \\ & = -3 &&& & = { 4 \pm \sqrt{32} \over 2} \\ \implies & \text{No real roots} &&& & = {4 \pm 4 \sqrt{2} \over 2} \\ & &&& & = 2 + 2 \sqrt{2} \text{ or } 2 - 2\sqrt{2} \end{align}

\begin{align} \log_{\sqrt{a}} \left(1 \over x\right) + \log_a x + \log_{a^2} x + \log_{a^4} x & = c \\ {\log_a \left(1 \over x\right) \over \log_a \sqrt{a}} + \log_a x + {\log_a x \over \log_a a^2} + {\log_a x \over \log_a a^4} & = c \phantom{0000000} [\text{Change-of-base}] \\ {\log_a x^{-1} \over \log_a a^{1 \over 2}} + \log_a x + {\log_a x \over 2} + {\log_a x \over 4} & = c \\ \\ {- \log_a x \over {1 \over 2}} + \log_a x + {\log_a x \over 2} + {\log_a x \over 4} & = c \\ - 2 \log_a x + \log_ax + {\log_a x \over 2} + {\log_a x \over 4} & = c \\ - \log_a x + {\log_a x \over 2} + {\log_a x \over 4} & = c \\ \\ 4 \left( - \log_a x + {\log_a x \over 2} + {\log_a x \over 4} \right) & = 4c \\ - 4 \log_a x + 2 \log_a x + \log_a x & = 4c \\ \\ - \log_a x & = 4c \\ \log_a x & = - 4c \\ x & = a^{-4c} \end{align}

\begin{align} 2 \log_a b + 4 \log_b a & = 9 \\ 2 \log_a b + 4 \left( \log_a a \over \log_a b \right) & = 9 \\ 2 \log_a b + 4 \left(1 \over \log_a b \right) & = 9 \\ \\ \text{Let } & u = \log_a b, \\ 2u + 4 \left(1 \over u\right) & = 9 \\ 2u + {4 \over u} & = 9 \\ 2u^2 + 4 & = 9u \\ 2u^2 - 9u + 4 & = 0 \\ (2u - 1)(u - 4) & = 0 \end{align} \begin{align} 2u - 1 & = 0 && \text{ or } & u - 4 & = 0 \\ 2u & = 1 &&& u & = 4 \\ u & = {1 \over 2} \\ \\ \log_a b & = {1 \over 2} &&& \log_a b & = 4 \\ b & = a^{1 \over 2} &&& b & = a^4 \text{ (Reject, since } a > b) \\ b & = \sqrt{a} \end{align}