\begin{align} \log_4 8 & = {\log_2 8 \over \log_2 4} \\ & = {\log_2 2^3 \over \log_2 2^2} \\ & = {3 \log_2 2 \over 2 \log_2 2 } \phantom{00000} [\text{Power law}] \\ & = {3 (1) \over 2 (1) } \\ & = {3 \over 2} \end{align}

\begin{align} \text{L.H.S} & = 2\log_a 2 + \log_a 10 - 3\log_a 3 - \log_a 5 \\ & = \log_a 2^2 + \log_a 10 - \log_a 3^3 - \log_a 5 \phantom{00000} [\text{Power law}] \\ & = \log_a 4 + \log_a 10 - \log_a 27 - \log_a 5 \\ & = \log_a (4 \times 10) - \log_a 27 - \log_a 5 \phantom{0000000} [\text{Product law}] \\ & = \log_a 40 - \log_a 27 - \log_a 5 \\ & = \log_a {40 \over 27} - \log_a 5 \phantom{000000000000000000} [\text{Quotient law}] \\ & = \log_a { {40 \over 27} \over 5} \\ & = \log_a {8 \over 27} \\ & = \log_a \left(2 \over 3\right)^3 \\ & = 3 \log_a {2 \over 3} \phantom{0} \\ & = \text{R.H.S} \end{align}

\begin{align} \require{cancel} \text{L.H.S} & = \log_2 27 \times \log_3 25 \times \log_5 16 \\ & = \log_2 27 \times {\log_2 25 \over \log_2 3} \times {\log_2 16 \over \log_2 5} \phantom{00000} [\text{Change-of-base}] \\ & = {\log_2 27 \times \log_2 25 \times \log_2 16 \over \log_2 3 \times \log_2 5} \\ & = {\log_2 3^3 \times \log_2 5^2 \times \log_2 2^4 \over \log_2 3 \times \log_2 5} \\ & = {3 \cancel{\log_2 3} \times 2 \cancel{\log_2 5} \times 4 \log_2 2 \over \cancel{\log_2 3} \times \cancel{\log_2 5} } \phantom{00000} [\text{Power law}] \\ & = 3 \times 2 \times 4 \log_2 2 \\ & = 6 \times 4(1) \\ & = 24 \\ & = \text{R.H.S} \end{align}

\begin{align} \log_9 81x^3 & = \log_9 81 + \log_9 x^3 \phantom{000000} [\text{Product law}] \\ & = \log_9 9^2 + \log_9 x^3 \\ & = 2 \log_9 9 + 3 \log_9 x \phantom{000000} [\text{Power law}] \\ & = 2(1) + 3p \\ & = 2 + 3p \end{align}

\begin{align} (\log_3 \sqrt{3x})^2 & = [ \log_3 (3x)^{1 \over 2} ]^2 \\ & = \left[ {1 \over 2} \log_3 (3x) \right]^2 \phantom{000000000} [\text{Power law}] \\ & = \left[ {1 \over 2} (\log_3 3 + \log_3 x) \right]^2 \phantom{000} [\text{Product law}] \\ & = \left[ {1 \over 2} (1 + \log_3 x) \right]^2 \phantom{000000} [\text{Need to find value of } \log_3 x]\\ \\ \\ \log_9 x & = p \\ {\log_3 x \over \log_3 9} & = p \phantom{000000} [\text{Change-of-base}] \\ {\log_3 x \over \log_3 3^2} & = p \\ {\log_3 x \over 2 \log_3 3} & = p \\ {\log_3 x \over 2(1)} & = p \\ {\log_3 x \over 2} & = p \\ \log_3 x & = 2p \\ \\ \\ \therefore (\log_3 \sqrt{3x})^2 & = \left[{1 \over 2}(1 + 2p)\right]^2 \\ & = \left(1 \over 2\right)^2 (1 + 2p)^2 \\ & = {1 \over 4}(1 + 2p)^2 \end{align}

\begin{align} \log_4 (\log_3 x) & = 1 \\ \log_3 x & = 4^1 \phantom{00000} [\text{Change to exponential form}] \\ \log_3 x & = 4 \\ x & = 3^4 \phantom{00000} [\text{Change to exponential form}] \\ x & = 81 \end{align}

\begin{align} \log_4 (x + 10) - \log_4 (x - 2) & = \log_4 (x + 3) \\ \log_4 {x + 10 \over x - 2} & = \log_4 (x + 3) \phantom{000000} [\text{Quotient law}] \\ \\ {x + 10 \over x - 2} & = x + 3 \\ x + 10 & = (x - 2)(x + 3) \\ x + 10 & = x^2 + 3x - 2x - 6 \\ x + 10 & = x^2 + x - 6 \\ 16 & = x^2 \\ \pm \sqrt{16} & = x \end{align}
\begin{align} x & = 4 && \text{ or } & x & = -4 \text{ (Reject, since } \log_4 -6 \text{ is undefined}) \end{align}

\begin{align} (\log_2 x)^2 & = 9 \\ \log_2 x & = \pm \sqrt{9} \\ \log_2 x & = \pm 3 \end{align}
\begin{align} \log_2 x & = 3 && \text{ or } & \log_2 x & = -3 \\ x & = 2^3 &&& x & = 2^{-3} \phantom{00000} [\text{Change to exponential form}] \\ x & = 8 &&& x & = {1 \over 8} \end{align}

\begin{align} \log_4 x^2 + \log_{16} x & = \log_3 27 \\ \log_4 x^2 + \log_{16} x & = \log_3 3^3 \\ 2 \log_4 x + \log_{16} x & = 3 \log_3 3 \phantom{000000} [\text{Power law}] \\ 2 \log_4 x + \log_{16} x & = 3(1) \\ 2 \log_4 x + {\log_4 x \over \log_4 16} & = 3 \phantom{000000} [\text{Change-of-base}] \\ 2 \log_4 x + {\log_4 x \over \log_4 4^2} & = 3 \\ 2 \log_4 x + {\log_4 x \over 2 \log_4 4} & = 3 \\ 2 \log_4 x + {\log_4 x \over 2 (1)} & = 3 \\ 2 \log_4 x + {\log_4 x \over 2} & = 3 \\ 2 \left( 2 \log_4 x + {\log_4 x \over 2} \right) & = 2(3) \\ 4 \log_4 x + \log_4 x & = 6 \\ 5 \log_4 x & = 6 \\ \log_4 x & = {6 \over 5} \\ x & = 4^{6 \over 5} \phantom{000000} [\text{Change to exponential form}] \\ x & = 5.2780 \\ x & \approx 5.28 \end{align}

\begin{align} \log_x 4 - 3 \log_2 x & = 5 \\ \log_x 4 - 3 \left( \log_x x \over \log_x 2 \right) & = 5 \phantom{0000000} [\text{Change-of-base}] \\ \log_x 4 - 3 \left(1 \over \log_x 2\right) & = 5 \\ \log_x 2^2 - {3 \over \log_x 2} & = 5 \\ 2 \log_x 2 - {3 \over \log_x 2} & = 5 \phantom{0000000} [\text{Power law}] \\ \\ \text{Let } u = \log_x 2, & \\ 2u - {3 \over u} & = 5 \\ u \left(2u - {3 \over u}\right) & = 5u \\ 2u^2 - 3 & = 5u \\ 2u^2 - 5u - 3 & = 0 \\ (u - 3)(2u + 1) & = 0 \end{align}
\begin{align} u - 3 & = 0 && \text{ or } & 2u + 1 & =0 \\ u & = 3 &&& 2u & = -1 \\ & &&& u & = -{1 \over 2} \\ \\ \log_x 2 & = 3 &&& \log_x 2 & = -{1 \over 2} \\ 2 & = x^3 &&& 2 & = x^{-{1 \over 2}} \phantom{000000} [\text{Exponential form}] \\ \sqrt[3]{2} & = x &&& 2 & = {1 \over \sqrt{x}} \\ 1.26 & \approx x &&& 2 \sqrt{x} & = 1 \\ & &&& \sqrt{x} & = {1 \over 2} \\ & &&& x & = \left(1 \over 2\right)^2 \\ & &&& x & = {1 \over 4} \end{align}

\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 \\ \\ \\ \therefore x & = 3, y = 5 \end{align}

\begin{align} \ln (2x + y) - 3x & = 1 \\ \ln (2x + y) & = 3x + 1 \\ \log_e (2x + y) & = 3x + 1 \\ 2x + y & = e^{3x + 1} \phantom{000000} [\text{Exponential form}] \\ y & = e^{3x+ 1} - 2x \end{align}

\begin{align} x - e^{1 - {1 \over 2}x} & = 0 \\ x & = e^{1 - {1 \over 2}x} \\ \ln x & = \ln e^{1 - {1 \over 2}x} \\ \ln x & = \left(1 - {1 \over 2}x\right) \ln e \phantom{000000} [\text{Power law}] \\ \ln x & = \left(1 - {1 \over 2}x\right)(1) \\ \ln x & = 1 - {1 \over 2}x \\ 2 \ln x & = 2 \left(1 - {1 \over 2}x\right) \\ 2 \ln x & = 2 - x \\ \underbrace{2 \ln x}_\text{Curve} & = \underbrace{-x + 2}_\text{Line} \\ \\ \text{Draw } & y = - x + 2 \\ \\ \text{Let } & x = 0, y = 2 \\ \\ \text{Let } & y = 0, 0 = - x + 2 \\ & \phantom{00000.} x = 2 \\ \\ \text{From sketch, } & \text{equation has only 1 solution} \end{align}

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

\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 exponential form}] \\ \\ t & = e^{13 \over 6} - 1 \\ & = 7.729 \phantom{.} 1 \\ & \approx 8 \end{align}

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