(a)

\begin{align} \ln 4 + \lg 6 & = 2.16444 \\ & \approx 2.16 \end{align}

(b)

\begin{align} 3 \lg 2 - \ln 2 & = 0.20994 \\ & \approx 0.210 \end{align}

(c)

\begin{align} {\ln 7.1 \over 2\lg 5} & = 1.40213 \\ & \approx 1.40 \end{align}

(d)

\begin{align} {\lg 9 - \ln 3 \over \ln (e^2 - 1)} & = -0.077844 \\ & \approx -0.0778 \end{align}

For the logarithmic expression $ \log_a b $ to be defined, the following two conditions must be fulfilled: \begin{align} 1) \phantom{0} b & > 0 \\ \\ 2) \phantom{0} a & > 0 \text{ and } a \ne 1 \end{align}

(a)

\begin{align} \text{When } & x = 0.5, \\ \log_x (5 - 2x) & = \log_{0.5} [5 - 2(0.5)] \\ & = \log_{0.5} 4 \\ \\ \text{Comparing } & \text{with } \log_a b, \\ a & = 0.5 > 0 \\ b & = 4 > 0 \\ \\ \therefore \log_{0.5} 4 & \text{ is defined} \end{align}

(b)

\begin{align} \text{When } & x = 3, \\ \log_x (5 - 2x) & = \log_3 [5 - 2(3)] \\ & = \log_3 (-1) \\ \\ \text{Comparing } & \text{with } \log_a b, \\ a & = 3 > 0 \\ b & = -1 < 0 \phantom{00000000} [ y \text{ must be positive}] \\ \\ \therefore \log_{3} (-1) & \text{ is not defined} \end{align}

(c)

\begin{align} \text{When } & x = 2.5, \\ \log_x (5 - 2x) & = \log_{2.5} [5 - 2(2.5)] \\ & = \log_{2.5} 0 \\ \\ \text{Comparing } & \text{with } \log_a b, \\ a & = 2.5 > 0 \\ b & = 0 \phantom{00000000} [y \text{ must be positive}] \\ \\ \therefore \log_{2.5} 0 & \text{ is not defined} \end{align}

(d)

\begin{align} \text{When } & x = 1, \\ \log_x (5 - 2x) & = \log_{1} [5 - 2(1)] \\ & = \log_1 3 \\ \\ \text{Comparing } & \text{with } \log_a b, \\ a & = 1 \phantom{00000000} [ a \text{ cannot be equals to } 1] \\ \\ \therefore \log_{1} 3 & \text{ is not defined} \end{align}

(e)

\begin{align} \text{When } & x = \sqrt{2}, \\ \log_x (5 - 2x) & = \log_{\sqrt{2}} (5 - 2\sqrt{2}) \\ & = \log_{\sqrt{2}} (2.17157) \\ \\ \text{Comparing } & \text{with } \log_a b, \\ a & = \sqrt{2} > 0 \\ b & = 2.17157 > 0 \\ \\ \therefore \log_{\sqrt{2}} (2.17157) & \text{ is defined} \end{align}

Note: \begin{align} \underbrace{y = a^x}_{\text{Index form}} \iff \underbrace{x = \log_a y}_{\text{Logarithmic form}} \end{align}

(a)

\begin{align} {1 \over 9} & = 3^{-2} \\ \\ \implies \log_3 {1 \over 9} & = -2 \end{align}

(b)

\begin{align} 5 & = 10^n \\ \\ \implies \log_{10} 5 & = n \\ \lg 5 & = n \end{align}

(c)

\begin{align} 4 & = e^x \\ \\ \implies \log_{e} 4 & = x \\ \ln 4 & = x \end{align}

(d)

\begin{align} p & = 2^x \\ \\ \implies \log_{2} p & = x \end{align}

(e)

\begin{align} y & = a^3 \\ \\ \implies \log_{a} y & = 3 \end{align}

Note: \begin{align} \underbrace{y = a^x}_{\text{Index form}} \iff \underbrace{x = \log_a y}_{\text{Logarithmic form}} \end{align}

(a)

\begin{align} \log_5 125 & = 3 \\ \\ \implies 125 & = 5^3 \end{align}

(b)

\begin{align} \lg 100 & = 2 \\ \log_{10} 100 & = 2 \\ \\ \implies 100 & = 10^2 \end{align}

(c)

\begin{align} \ln x & = 2 \\ \log_e x & = 2 \\ \\ \implies x & = e^2 \end{align}

(d)

\begin{align} \log_x 3 & = 4 \\ \\ \implies 3 & = x^4 \end{align}

(e)

\begin{align} \log_3 y & = n \\ \\ \implies y & = 3^n \end{align}

Note: \begin{align} \underbrace{y = a^x}_{\text{Index form}} \iff \underbrace{x = \log_a y}_{\text{Logarithmic form}} \end{align}

(a)

\begin{align} \log_4 x & = 2 &\text{or}\phantom{0000} \log_2 y & = 3 \\ \\ \implies x & = 4^2 & y & = 2^3 \\ & = 16 & & = 8 \end{align} \begin{align} {x \over y} & = {16 \over 8} \\ & = 2 \end{align}

(b)

\begin{align} \log_x a & = 1 &\text{or}\phantom{0000} \log_y b & = 2 \\ \\ \implies a & = x^1 & b & = y^2 \\ a & = x \end{align} \begin{align} xy^2 & = (x)(y^2) \\ & = (a)(b) \\ & = ab \end{align}

Special results: \begin{align} \text{For } a > 0 &\text{ and } a \ne 1, \\ 1) \phantom{0} \log_a a & = 1 \\ 2) \phantom{0} \log_a 1 & = 0 \end{align}

(a)

\begin{align} \log_4 4 - 3\log_2 2 & = (1) - 3(1) \\ & = 1 - 3 \\ & = -2 \end{align}

(b)

\begin{align} \log_2 1 + 4\log_5 5 & = (0) + 4(1) \\ & = 0 + 4 \\ & = 4 \end{align}

(c)

\begin{align} (3 - \log_3 3)^3 & = [3 - (1)]^3 \\ & = 2^3 \\ & = 8 \end{align}

(d)

\begin{align} \left( 3\log_x x + 2 \over 4 - 2\log_5 1\right)^2 & = \left[ 3(1) + 2 \over 4 - 2(0) \right]^2 \\ & = \left(3 + 2 \over 4\right)^2 \\ & = \left( 5 \over 4 \right)^2 \\ & = {25 \over 16} \end{align}

(e)

\begin{align} \log_2 (6 - 5\log_7 7) & = \log_2 [6 - 5(1)] \\ & = \log_2 (6 - 5) \\ & = \log_2 1 \\ & = 0 \end{align}

(f)

\begin{align} \log_2 (4 - 2\lg 10) & = \log_2 (4 - 2\log_{10} 10) \\ & = \log_2 [4 - 2(1)] \\ & = \log_2 (4 - 2) \\ & = \log_2 2 \\ & = 1 \end{align}

Note: \begin{align} \underbrace{y = a^x}_{\text{Index form}} \iff \underbrace{x = \log_a y}_{\text{Logarithmic form}} \end{align}

(a)

\begin{align} \log_2 x & = 3 \\ \\ \implies x & = 2^3 \\ & = 8 \end{align}

(b)

\begin{align} \log_x 9 & = 2 \\ \\ \implies 9 & = x^2 \\ \\ x & = \pm\sqrt{9} \\ & = \pm 3 \\ & = 3 \phantom{0} \text{ or} \phantom{0} -3 \text{ (Reject, since } \log_{-3} 9 \text{ is undefined}) \\ \\ \therefore x & = 3 \end{align}

(c)

\begin{align} \log_4 8 & = x \\ \\ \implies 8 & = 4^x \\ 2^3 & = (2^2)^x \\ 2^3 & = 2^{2x} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ \\ \therefore 3 & = 2x \\ {3 \over 2} & = x \end{align}

Note: \begin{align} \underbrace{y = a^x}_{\text{Index form}} \iff \underbrace{x = \log_a y}_{\text{Logarithmic form}} \end{align}

(a)

\begin{align} \lg x & = 0.61 \\ \log_{10} x & = 0.61 \\ \\ \implies x & = 10^{0.61} \\ & = 4.0738 \\ & \approx 4.07 \end{align}

(b)

\begin{align} (\ln x)^2 & = 3 \\ \ln x & = \pm\sqrt{3} \end{align} \begin{align} \ln x & = \sqrt{3} & \ln x & = -\sqrt{3} \\ \log_e x & = \sqrt{3} & \log_e x & = -\sqrt{3} \\ \\ \implies x & = e^{\sqrt{3}} & x & = e^{-\sqrt{3}} \\ & = 5.6522 & & = 0.17692 \\ & \approx 5.65 & & \approx 0.177 \end{align}

(c)

\begin{align} \ln x & = \lg 2 \\ \log_e x & = \lg 2 \\ \\ \implies x & = e^{\lg 2} \\ & = 1.3512 \\ & \approx 1.35 \end{align}

(d)

\begin{align} \lg (3x) & = 9 \\ \log_{10} (3x) & = 9 \\ \\ \implies 3x & = 10^9 \\ & = 1000\phantom{.}000\phantom{.}000 \\ x & = 333333333.333 \\ & = 3.3333333333 \times 10^8 \\ & \approx 3.33 \times 10^8 \end{align}

(a)

\begin{align} k & = e^{2x} \\ \\ \implies \log_e k & = 2x \\ \ln k & = 2x \end{align}

(b)

\begin{align} 9 & = 10^{x - 4} \\ \\ \implies \log_{10} 9 & = x - 4 \\ \lg 9 & = x - 4 \end{align}

(c)

\begin{align} 2 - k & = x^4 \\ (2 - k) & = x^4 \\ \\ \implies \log_x (2 - k) & = 4 \\ \end{align}

(d)

\begin{align} x - 2 & = e^{m + 5} \\ (x - 2) & = e^{m + 5} \\ \\ \implies \log_e (x - 2) & = m + 5 \\ \ln (x - 2) & = m + 5 \end{align}

(a)

\begin{align} \log_3 y & = n + 1 \\ \log_3 y & = (n + 1) \\ \\ \implies y & = 3^{n + 1} \end{align}

(b)

\begin{align} \ln k & = x - 3 \\ \log_e k & = (x - 3) \\ \\ \implies k & = e^{x - 3} \end{align}

(c)

\begin{align} \lg (x - y) & = \sqrt{2} \\ \log_{10} (x - y) & = \sqrt{2} \\ \\ \implies x - y & = 10^{\sqrt{2}} \end{align}

(d)

\begin{align} \log_2 (4y) & = p + 1 \\ \log_2 (4y) & = (p + 1) \\ \\ \implies 4y & = 2^{p + 1} \end{align}

\begin{align} \log_4 y & = a \\ \\ \implies y & = 4^a \phantom{0} \text{ --- (1)} \\ \\ \log_8 (2y) & = b \\ \\ \implies 2y & = 8^b \phantom{0} \text{ --- (2)} \\ \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(4^a) & = 8^b \\ 2[(2^2)^a] & = (2^3)^b \\ 2(2^{2a}) & = 2^{3b} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ (2^1)(2^{2a}) & = 2^{3b} \\ [ (2^m)(2^n) = 2^{m + n} ] \phantom{00000000} 2^{1 + 2a} & = 2^{3b} \\ \\ \therefore 1 + 2a & = 3b \\ 2a & = 3b - 1 \text{ (Shown)} \end{align}

\begin{align} \log_3 p & = a \\ \\ \implies p & = 3^a \phantom{0} \text{ --- (1)} \\ \\ \log_{27} q & = b \\ \\ \implies q & = 27^b \phantom{0} \text{ --- (2)} \\ \\ \\ {p \over q} & = 3^c \\ \\ \text{Substitute } & \text{(1) and (2),} \\ {3^a \over 27^b} & = 3^c \\ {3^a \over (3^3)^b} & = 3^c \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {3^a \over 3^{3b}} & = 3^c \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{0000000} 3^{a - 3b} & = 3^c \\ \\ \therefore a - 3b & = c \end{align}

(a)

\begin{align} \log_2 (2x + 1) & = -3 \\ \\ \implies 2x + 1 & = 2^{-3} \\ 2x + 1 & = {1 \over 8} \\ 2x & = -{7 \over 8} \\ x & = -{7 \over 16} \end{align}

(b)

\begin{align} \log_3 (x^2 - 1) & = 1 \\ \\ \implies x^2 - 1 & = 3^1 \\ x^2 - 1 & = 3 \\ x^2 & = 4 \\ x & = \pm\sqrt{4} \\ & = \pm 2 \end{align}

(c)

\begin{align} \log_x (6x - 8) & = 2 \\ \\ \implies 6x - 8 & = x^2 \\ 0 & = x^2 - 6x + 8 \\ 0 & = (x - 4)(x - 2) \end{align} \begin{align} x - 4 & = 0 &\text{or}\phantom{0000} x - 2 & = 0 \\ x & = 4 & x & = 2 \end{align}

(d)

\begin{align} \log_x 64 & = {3 \over 2} \\ \\ \implies 64 & = x^{3 \over 2} \\ (64)^2 & = (x^{3 \over 2})^2 \\ 4096 & = x^3 \\ \sqrt[3]{4096} & = x \\ 16 & = x \end{align}

(e)

\begin{align} \log_9 \sqrt{27} & = x + 1 \\ \\ \implies \sqrt{27} & = 9^{x + 1} \\ 27^{1 \over 2} & = 9^{x + 1} \\ (3^3)^{1 \over 2} & = (3^2)^{x + 1} \\ 3^{3 \over 2} & = 3^{2(x + 1)} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ \\ \therefore {3 \over 2} & = 2(x + 1) \\ {3 \over 2} & = 2x + 2 \\ {3 \over 2} - 2 & = 2x \\ -{1 \over 2} & = 2x \\ -{1 \over 2} \div 2 & = x \\ -{1 \over 4} & = x \end{align}

(a)

\begin{align} \ln 2 \times \ln (4x) & = 3 \\ \ln (4x) & = {3 \over \ln 2} \\ \ln (4x) & = 4.32808 \\ \log_e (4x) & = 4.32808 \\ \\ \implies 4x & = e^{4.32808} \\ 4x & = 75.799 \\ x & = 75.799 \div 4 \\ & = 18.949 \\ & \approx 18.9 \end{align}

(b)

\begin{align} \lg (x - 2) & = (\lg 3)^2 \\ \lg (x - 2) & = 0.22764 \\ \log_{10} (x - 2) & = 0.22764 \\ \\ \implies x - 2 & = 10^{0.22764} \\ x & = 10^{0.22764} + 2 \\ & = 3.6890 \\ & \approx 3.69 \end{align}

(c)

\begin{align} \ln (4x) & = \lg 3 \times \lg 5 \\ \ln (4x) & = 0.33349 \\ \log_e (4x) & = 0.33349 \\ \\ \implies 4x & = e^{0.33349} \\ x & = e^{0.33349} \div 4 \\ & = 0.34895 \\ & \approx 0.349 \end{align}

(d)

\begin{align} \lg (x - 1) & = \lg (e^2 - 1) \\ \lg (x - 1) & = 0.80543 \\ \log_{10} (x - 1) & = 0.80543 \\ \\ \implies x - 1 & = 10^{0.80543} \\ x & = 10^{0.80543} + 1 \\ & = 7.3890 \\ & \approx 7.39 \end{align}

(a)

\begin{align} \log_x 16 & = 4 \\ \\ \implies 16 & = x^4 \\ 16 & = x^4 \\ \\ x & = \pm \sqrt[4]{16} \\ & = \pm 2 \\ & = 2 \text{ or } - 2 \text{ (Reject, since } \log_{-2} 16 \text{ is undefined}) \\ \\ \\ \log_2 y & = x \\ \\ \implies y & = 2^x \\ & = 2^2 \\ & = 4 \end{align}

(b)

\begin{align} \log_y x & = 2 \\ \\ \implies x & = y^2 \phantom{000} \text{ --- (1)} \\ \\ xy & = 8 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ (y^2)y & = 8 \\ y^3 & = 8 \\ y & = \sqrt[3]{8} \\ & = 2 \\ \\ \text{Substitute } & y = 2 \text{ into (1),} \\ x & = (2)^2 \\ & = 4 \end{align}

\begin{align} \log_2 (\log_3 x) & = \ln e \\ \log_2 (\log_3 x) & = \log_e e \\ \log_2 (\log_3 x) & = 1 \phantom{00000000} [\log_a a = 1] \\ \\ \implies \log_3 x & = 2^1 \phantom{0000000} [\text{Change to index form}] \\ \log_3 x & = 2 \\ \\ \implies x & = 3^2 \phantom{0000000} [\text{Change to index form}] \\ & = 9 \end{align}

For the logarithmic expression $ \log_a b $ to be defined, the following two conditions must be fulfilled: \begin{align} 1) \phantom{0} b & > 0 \\ \\ 2) \phantom{0} a & > 0 \text{ and } a \ne 1 \end{align}

(a)

\begin{align} \lg (x + 2) & = \log_{10} (x + 2) \\ \\ x + 2 & > 0 \\ x & > - 2 \end{align}

(b)

\begin{align} \ln (x^2 - 2x) & = \log_e (x^2 - 2x) \\ \\ x^2 - 2x & > 0 \\ x(x - 2) & > 0 \end{align}

$$ x < 0 \text{ or } x > 2 $$

(c)

\begin{align} \text{1st condition: } 3 - x & > 0 \phantom{00} & \text{2nd condition: } x & > 0, x \ne 1 \\ -x & > - 3 \\ x & < 3 \\ \\ \therefore 0 < x < 1 & \text{ or } 1 < x < 3 \end{align}

(a)

\begin{align} \ln (y + 1) - x & = 0 \\ \ln (y + 1) & = x \\ \log_e (y + 1) & = x \\ \\ \implies y + 1 & = e^x \\ y & = e^x - 1 \end{align}

(b)

\begin{align} 2\lg y & = x - 2 \\ \lg y & = {1 \over 2}x - 1 \\ \log_{10} y & = {1 \over 2}x - 1 \\ \\ \implies y & = 10^{{1 \over 2}x - 1} \end{align}

(c)

\begin{align} e^{2y} + 4 & = x \\ e^{2y} & = x - 4 \\ \\ x - 4 & = e^{2y} \\ \\ \implies \log_e (x - 4) & = 2y \\ \ln (x - 4) & = 2y \\ \\ 2y & = \ln (x - 4) \\ y & = {1 \over 2}\ln (x - 4) \end{align}

(d)

\begin{align} \ln (x + y) - 4x & = 0 \\ \ln (x + y) & = 4x \\ \log_e (x + y) & = 4x \\ \\ \implies x + y & = e^{4x} \\ y & = e^{4x} - x \end{align}

\begin{align} \log_e (x^2 + 1 - e^{3\lg 10}) & = 3 \\ \implies \\ x^2 + 1 - e^{3\lg 10} & = e^3 \\ [\lg 10 = 1] \phantom{00000000} x^2 + 1 - e^{3(1)} & = e^3 \\ x^2 + 1 - e^3 & = e^3 \\ x^2 & = e^3 + e^3 - 1 \\ x^2 & = 2e^3 - 1 \\ x & = \pm \sqrt{2e^3 - 1} \\ & = \pm 6.2586 \\ & \approx \pm 6.26 \end{align}

(i)

\begin{align} \log_2 \sqrt{8} & = p \\ \\ \implies \sqrt{8} & = 2^p \phantom{00000000} [\text{Change to index form}] \\ (8)^{1 \over 2} & = 2^p \\ (2^3)^{1 \over 2} & = 2^p \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^{3 \times {1 \over 2}} & = 2^p \\ 2^{3 \over 2} & = 2^p \\ \\ \therefore p & = {3 \over 2} \end{align}

(ii)

\begin{align} \log_9 (31 + \log_x 16) & = \underbrace{\log_2 \sqrt{8}}_{= p} \\ \log_9 (31 + \log_x 16) & = {3 \over 2} \\ \\ \implies 31 + \log_x 16 & = 9^{3 \over 2} \\ 31 + \log_x 16 & = (3^2)^{3 \over 2} \\ 31 + \log_x 16 & = 3^{2({3 \over 2})} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ 31 + \log_x 16 & = 3^3 \\ 31 + \log_x 16 & = 27 \\ \log_x 16 & = 27 - 31 \\ \log_x 16 & = -4 \\ \\ \implies 16 & = x^{-4} \phantom{00000000} [\text{Change to index form}] \\ 16 & = {1 \over x^4} \\ x^4 & = {1 \over 16} \\ x & = \pm \sqrt[4]{1 \over 16} \\ & = \pm {1 \over 2} \\ & = {1 \over 2} \text{ or } -{1 \over 2} \text{ (Reject, since } \log_{-{1 \over 2}} 16 \text{ is undefined}) \end{align}

\begin{align} x^2 - 4x + \log_2 p & = 0 \\ \\ [a = 1, \phantom{0} b = - 4, \phantom{0} c & = \log_2 p] \\ \\ b^2 - 4ac & = (-4)^2 - 4(1)(\log_2 p) \\ & = 16 - 4\log_2 p \\ \\ b^2 - 4ac & = 0 \phantom{00000000} [\text{Equal real roots or 1 real root}] \\ 16 - 4\log_2 p & = 0 \\ -4 \log_2 p & = -16 \\ \log_2 p & = {-16 \over -4} \\ \log_2 p & = 4 \\ \\ \implies p & = 2^4 \\ & = 16 \end{align}

(i) Common logarithms refers to the logarithm with base 10

\begin{align} & \text{(a) } \lg 546 \approx 2.74 \\ & \text{(b) } \lg 12 \phantom{.} 458 \approx 4.10 \\ & \text{(c) } \lg 464 \phantom{.} 777 \phantom{.} 399 \approx 8.67 \end{align}

(ii)

\begin{align} & \text{If } \lg k \text{ is an integer, no. of digits} = \lg k + 1 \\ \\ & \text{Otherwise, no. of digits} = \text{Round down value of } \lg k + 1 \end{align}

(iii)

\begin{align} \lg 3^{42} & = 20.039 \\ \\ \text{No. of digits in } 3^{42} & = 20 + 1 \\ & = 21 \\ \\ \\ \lg 7^{32} & = 27.043 \\ \\ \text{No. of digits in } 7^{32} & = 27 + 1 \\ & = 28 \end{align}

(iv)

\begin{align} & 10^0 = 1 \phantom{000} \log 10^0 = 0 \\ & 10^1 = 10 \phantom{000} \log 10^1 = 1 \\ \\ & \text{If } x \text{ is a 1 digit integer, i.e. } 0 \le x \le 9, \text{ then } 0 \le \lg x < 1 \\ \\ \\ & 10^1 = 10 \phantom{000} \log 10^1 = 1 \\ & 10^2 = 100 \phantom{000} \log 10^2 = 2 \\ \\ & \text{If } x \text{ is a 2 digit integer, i.e. } 10 \le x \le 99, \text{ then } 1 \le \lg x < 2 \\ \\ \\ & \text{Thus, in general, no. of digits} = \text{Round down value of } \lg x + 1 \end{align}