\begin{align} {9^x \over 3^{x + 1}} & = 5^{2 - x} \\ {(3^2)^x \over 3^{x + 1}} & = 5^{2 - x} \\ {3^{2x} \over 3^{x + 1}} & = 5^{2 - x} \phantom{000000.} [ (a^m)^n = a^{mn}] \\ 3^{2x - (x + 1)} & = 5^{2 - x} \phantom{000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ 3^{x - 1} & = 5^{2 - x} \\ {3^x \over 3^1} & = {5^2 \over 5^x} \\ (3^x)(5^x) & = (3)(5^2) \\ 15^x & = 75 \phantom{00000000} [a^m \times b^m = (a \times b)^m] \end{align}

\begin{align} \require{cancel} { 7^{3x + 1} \times 49^{-1} \over 7^{3x + 1} - 343^x } & = { 7^{3x + 1} \times (7^2)^{-1} \over 7^{3x + 1} - 7^{3x}} \\ & = { 7^{3x + 1} \times 7^{-2} \over 7^{3x} \times 7 - 7^{3x} } \\ & = { 7^{3x + 1 + (-2)} \over 7^{3x} (7 - 1) } \phantom{000000} [\text{Factorise}] \\ & = { 7^{3x - 1} \over 7^{3x} (6) } \\ & = { \cancel{7^{3x}} \times 7^{-1} \over \cancel{7^{3x}} (6)} \\ & = { 7^{-1} \over 6 } \\ & = {1 \over 42} \end{align}

\begin{align} 4^{n + 1} + 3(4^{n + 2}) - 2^{2n + 1} & = 4^n \times 4 + 3(4^n \times 4^2) - 2^{2n} \times 2 \\ & = 4(4^n) + 3[16(4^n)] - 2(2^{2n}) \\ & = 4(4^n) + 48(4^n) - 2[ (2^2)^n] \\ & = 4(4^n) + 48(4^n) - 2(4^n) \\ & = 50(4^n) \\ & = 10[5 (4^n)] \\ \\ \therefore \text{Expression } & \text{is a multiple of } 10 \end{align}

\begin{align} 3^{2x + 1} - 3^{2x} & = 54 \\ (3^{2x})(3) - 3^{2x} & = 54 \\ 3^{2x} (3 - 1) & = 54 \phantom{000000} [\text{Factorise } 3^{2x}] \\ 3^{2x} (2) & = 54 \\ 3^{2x} & = {54 \over 2} \\ 3^{2x} & = 27 \\ 3^{2x} & = 3^3 \\ \\ 2x & = 3 \\ x & = {3 \over 2} \end{align}

\begin{align} 4 e^x - 5e^{1 - x} & = 0 \\ 4 e^x & = 5 e^{1 - x} \\ {e^x \over e^{1 - x}} & = {5 \over 4} \\ e^{x - (1 - x)} & = {5 \over 4} \\ e^{x - 1 + x} & = {5 \over 4} \\ e^{2x - 1} & = {5 \over 4} \\ \ln e^{2x - 1} & = \ln {5 \over 4} \\ (2x - 1)\ln e & = \ln {5 \over 4} \phantom{000000} [\text{Power law (logarithms)}] \\ (2x - 1)(1) & = \ln {5 \over 4} \\ 2x - 1 & = \ln {5 \over 4} \\ 2x & = \ln {5 \over 4} + 1 \\ x & = {1 \over 2} \left( \ln {5 \over 4} + 1 \right) \\ & \approx 0.612 \end{align}

\begin{align} 2^x + 2^{3 - x} & = 9 \\ 2^x + {2^3 \over 2^x} & = 9 \\ \\ \text{Let } u = 2^x, & \\ u + {8 \over u} & = 9 \\ u \left(u + {8 \over u}\right) & = 9u \\ u^2 + 8 & = 9u \\ u^2 - 9u + 8 & = 0 \\ (u - 1)(u - 8) & = 0 \end{align}
\begin{align} u - 1 & = 0 && \text{ or } & u - 8 & = 0 \\ u & = 1 &&& u & = 8 \\ \\ 2^x & = 1 &&& 2^x & = 8 \\ 2^x & = 2^0 &&& 2^x & = 2^3 \\ x & = 0 &&& x & = 3 \end{align}

\begin{align} 9^x - 4(3^x) - 5 & = 0 \\ (3^2)^x - 4(3^x) - 5 & = 0 \\ (3^x)^2 - 4(3^x) - 5 & = 0 \\ \\ \text{Let } & u = 3^x, \\ u^2 - 4u - 5 & = 0 \\ (u - 5)(u + 1) & = 0 \\ \\ u - 5 = 0 \phantom{00} & \text{ or } \phantom{00} u + 1 =0 \\ u = 5 \phantom{00} & \phantom{00000000} u = -1 \\ \\ \text{Since } & u = 3^x, \\ 3^x = 5 \phantom{0000.} & \text{ or } \phantom{00000} 3^x = -1 \text{ (Reject, since } 3^x > 0) \\ \lg 3^x = \lg 5 \phantom{00.} & \\ x\lg 3 = \lg 5 \phantom{00.} & \\ x = {\lg 5 \over \lg 3} \phantom{0(} & \\ x = 1.4549 & \\ x \approx 1.45 \phantom{00} & \\ \\ \\ \therefore x = 1.45 & \text{ is the only solution} \end{align}

\begin{align} {9^{2x} \over 81^{y - 1}} & = 27 \\ {(3^2)^{2x} \over (3^4)^{y - 1}} & = 3^3 \\ {3^{4x} \over 3^{4y - 4} } & = 3^3 \\ 3^{4x - (4y - 4)} & = 3^3 \\ \\ 4x - (4y - 4) & = 3 \\ 4x - 4y + 4 & = 3 \\ 4x - 4y & = -1 \phantom{00} \text{--- (1)} \\ \\ \\ 4^{3y} \times 8 & = 32^x \\ (2^2)^{3y} \times 2^3 & = (2^5)^x \\ 2^{6y} \times 2^3 & = 2^{5x} \\ 2^{6y + 3} & = 2^{5x} \\ \\ 6y + 3 & = 5x \\ 6y & = 5x - 3 \\ y & = {1 \over 6}(5x - 3) \\ y & = {5 \over 6}x - {1 \over 2} \phantom{00} \text{--- (2)} \\ \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 4x - 4\left({5 \over 6}x - {1 \over 2}\right) & = -1 \\ 4x - {10 \over 3}x + 2 & = -1 \\ {2 \over 3}x & = -3 \\ x & = -3 \div {2 \over 3} \\ x & = -{9 \over 2} \\ \\ \text{Substitute } & \text{into (2),} \\ y & = {5 \over 6} \left(-{9 \over 2}\right) - {1 \over 2} \\ y & = -{17 \over 4} \\ \\ \\ \therefore x & = -{9 \over 2}, y = -{17 \over 4} \end{align}

\begin{align} 2.5^x + x & = 6 \\ \underbrace{2.5^x}_{\text{Curve }} & = \underbrace{6 - x}_{\text{Line}} \\ \\ \text{Draw } & y = 6 - x \\ \\ \text{Let } & x = 0, y = 6 \\ \\ \text{Let } & y = 0, x = 6 \\ \\ \\ \text{From sketch, } & \text{there is only 1 solution} \end{align}

\begin{align} y & = ae^{bx} \\ \\ \text{Using } (0.5 , 1.1), & \text{ when } x = 0.5, y = 1.1, \\ 1.1 & = ae^{b(0.5)} \\ 1.1 & = ae^{0.5b} \phantom{000} \text{--- (1)} \\ \\ \text{Using } (1.5, 0.15), & \text{ when } x = 1.5, y = 0.15, \\ 0.15 & = ae^{b(1.5)} \\ 0.15 & = ae^{1.5b} \phantom{000} \text{--- (2)} \\ \\ (2) & \div (1), \\ {0.15 \over 1.1} & = {ae^{1.5b} \over ae^{0.5b}} \\ {3 \over 22} & = {e^{1.5b} \over e^{0.5b}} \\ {3 \over 22} & = e^{1.5b - 0.5b} \\ {3 \over 22} & = e^b \\ \\ \text{Taking } \ln & \text{ of both sides,} \\ \ln {3 \over 22} & = \ln e^b \\ \ln {3 \over 22} & = b \ln e \\ \ln {3 \over 22} & = b (1) \\ \ln {3 \over 22} & = b \\ \\ \text{Substitute } & b = \ln {3 \over 22} \text{ into (1),} \\ 1.1 & = ae^{0.5 \ln {3 \over 22}} \\ 1.1 & = a(0.36927) \\ {1.1 \over 0.36927} & = a \\ 2.9788 & = a \\ 2.98 & \approx a \\ \\ \therefore a & \approx 2.98, b = \ln {3 \over 22} = -1.9924 \approx -1.99 \end{align}

\begin{align} \text{When } & t = 0, \\ P & = 600(2 + e^{-0.2(0)}) \\ & = 600(2 + 1) \\ & = 1 \phantom{.} 800 \end{align}

\begin{align} \text{When } & t = 12, \\ P & = 600(2 + e^{-0.2(12)}) \\ & = 1 \phantom{.} 254.430 \phantom{.} 77 \\ & \approx 1 \phantom{.} 254 \end{align}

\begin{align} P & = 600(2 + e^{-0.2t}) \\ & = 600\left( 2 + {1 \over e^{0.2t}} \right) \end{align} \begin{align} \text{As } t \rightarrow \infty &, \phantom{0} {1 \over e^{0.2t}} \rightarrow 0 \\ \\ \therefore P & = 600(2 + 0) \\ & = 1 \phantom{.} 200 \end{align}

\begin{align} \text{Value after 1 year, } V & = 10 \phantom{.} 000 \times {105 \over 100} \phantom{000000} [100\% + 5\% = 105\%] \\ & = 10 \phantom{.} 000 (1.05) \\ \\ \text{Value after 2 years, } V & = 10 \phantom{.} 000 (1.05) \times {105 \over 100} \phantom{000000} [105\% \text{ of the amount from year 1}] \\ & = 10 \phantom{.} 000 (1.05)^2 \\ \\ \text{Value after 3 years, } V & = 10 \phantom{.} 000 (1.05)^2 \times {105 \over 100} \phantom{000000} [105\% \text{ of the amount from year 2}] \\ & = 10 \phantom{.} 000 (1.05)^3 \\ & . \\ & . \\ & . \\ \text{Value after } n \text{ years, } V & = 10 \phantom{.} 000 (1.05)^n \phantom{00} (\text{Shown}) \end{align}

\begin{align} \text{Initial value} & = \$10 \phantom{.} 000 \\ \\ \text{Triple of initial value} & = \$ 30 \phantom{.} 000 \\ \\ \text{Let } & V = 30 \phantom{.} 000, \\ 30 \phantom{.} 000 & = 10 \phantom{.} 000 (1.05)^n \\ {30 \phantom{.} 000 \over 10 \phantom{.} 000} & = (1.05)^n \\ 3 & = (1.05)^n \\ \lg 3 & = \lg (1.05)^n \\ \lg 3 & = n \lg 1.05 \phantom{000000} [\text{Power law (logarithms)}] \\ {\lg 3 \over \lg 1.05} & = n \\ 22.517 & = n \\ \\ \text{Number of years} & = 23 \end{align}