(a)

\begin{align} 3^{2x} & = 27 \\ 3^{2x} & = 3^3 \\ \\ \therefore 2x & = 3 \\ x & = {3 \over 2} \end{align}

(b)

\begin{align} 4^x & = 32 \\ (2^2)^x & = 2^5 \\ [(a^m)^n = a^{mn}] \phantom{00000000} 2^{2x} & = 2^5 \\ \\ \therefore 2x & = 5 \\ x & = {5 \over 2} \end{align}

(c)

\begin{align} (\sqrt{2})^{3x} & = {1 \over 8} \\ (2^{1 \over 2})^{3x} & = {1 \over 8} \\ [ (a^m)^n = a^{mn}] \phantom{00000000} 2^{3x \over 2} & = {1 \over 8} \\ 2^{3x \over 2} & = {1 \over 2^3} \\ 2^{3x \over 2} & = 2^{-3} \phantom{00000000} \left[ {1 \over a^m} = a^{-m} \right] \\ \\ \therefore {3x \over 2} & = -3 \\ 3x & = 2(-3) \\ & = -6 \\ x & = {-6 \over 3} \\ & = -2 \end{align}

(d)

\begin{align} \left(1 \over 9\right)^{x + 2} & = 3 \\ \left(1 \over 3^2 \right)^{x + 2} & = 3 \\ \left[ {1 \over a^m} = a^{-m} \right] \phantom{00000000} (3^{-2})^{x + 2} & = 3 \\ \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 3^{-2(x + 2)} & = 3 \\ 3^{-2x - 4} & = 3^1 \\ \\ \therefore -2x - 4 & = 1 \\ -2x & = 1 + 4 \\ & = 5 \\ x & = -{5 \over 2} \end{align}

(a)

\begin{align} 4^x(5^{2x}) & = 10 \\ (2^2)^x (5^{2x}) & = 10 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^{2x} (5^{2x}) & = 10 \\ [ a^m \times b^m = (a \times b)^m ] \phantom{0000000} (2 \times 5)^{2x} & = 10 \\ 10^{2x} & = 10^1 \\ \\ \therefore 2x & = 1 \\ x & = {1 \over 2} \end{align}

(b)

\begin{align} {4^x \over 2^{x - 1}} & = 8^{2 - x} \\ {(2^2)^x \over 2^{x - 1}} & = (2^3)^{2 - x} \\ {2^{2x} \over 2^{x - 1}} & = 2^{3(2 - x)} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 2^{2x - (x - 1)} & = 2^{3(2 - x)} \\ \\ \therefore 2x - (x - 1) & = 3(2 - x) \\ 2x - x + 1 & = 6 - 3x \\ x + 1 & = 6 - 3x \\ x + 3x & = 6 - 1 \\ 4x & = 5 \\ x & = {5 \over 4} \end{align}

(c)

\begin{align} 4^{x^2 - 6} - 16^{x + 1} & = 0 \\ 4^{x^2 - 6} & = 16^{x + 1} \\ 4^{x^2 - 6} & = (4^2)^{x + 1} \\ 4^{x^2 - 6} & = 4^{2(x + 1)} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ \\ \therefore x^2 - 6 & = 2(x + 1) \\ x^2 - 6 & = 2x + 2 \\ x^2 - 2x - 6 - 2 & = 0 \\ x^2 - 2x - 8 & = 0 \\ (x - 4)(x + 2) & = 0 \\ \\ x - 4 = 0 \phantom{00} &\text{or} \phantom{00} x + 2 = 0 \\ x = 4 \phantom{00} & \phantom{or00+2} x = - 2 \end{align}

(d)

\begin{align} 3^{x^2} & = {9^{x + 1} \over 27^x} \\ 3^{x^2} & = {(3^2)^{x + 1} \over (3^3)^x} \\ 3^{x^2} & = {3^{2(x + 1)} \over 3^{3x}} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ 3^{x^2} & = 3^{2(x + 1) - 3x} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \therefore x^2 & = 2(x + 1) - 3x \\ x^2 & = 2x + 2 - 3x \\ x^2 & = -x + 2 \\ x^2 + x - 2 & = 0 \\ (x + 2)(x - 1) & = 0 \\ \\ x + 2 = 0 \phantom{000}&\text{or} \phantom{00} x - 1 = 0 \\ x = -2 \phantom{0.}& \phantom{or00-1} x = 1 \end{align}

(i)

\begin{align} 3(2^{x - 1}) & = 2^x + 4 \\ \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{00000000} 3 \left( 2^x \over 2^1 \right) & = 2^x + 4 \\ 3 \left( 2^x \over 2 \right) & = 2^x + 4 \\ {3 \over 2}(2^x) & = 2^x + 4 \\ \\ \text{When } & u = 2^x, \\ {3 \over 2}u & = u + 4 \end{align}

(ii)

\begin{align} {3 \over 2}u & = u + 4 \\ {3 \over 2}u - u & = 4 \\ {1 \over 2}u & = 4 \\ u & = 8 \\ \\ \text{Since } & u = 2^x, \\ 2^x & = 8 \\ 2^x & = 2^3 \\ \\ \therefore x & = 3 \end{align}

\begin{align} 3^x & = 4 - 3(3^{-x}) \\ 3^x & = 4 - 3\left(1 \over 3^x\right) \phantom{00000000} \left[ {1 \over a^m} = a^{-m} \right] \\ 3^x & = 4 - {3 \over 3^x} \\ \\ \text{When } & u = 3^x, \\ u & = 4 - {3 \over u} \\ u^2 & = 4u - 3 \\ u^2 - 4u + 3 & = 0 \\ (u - 3)(u - 1) & = 0 \\ \\ u - 3 = 0 \phantom{00}&\text{or}\phantom{00} u - 1 = 0 \\ u = 3 \phantom{00}&\phantom{or00-1} u = 1 \\ \\ \text{Since } & u = 3^x, \\ 3^x = 3 \phantom{00}&\text{or} \phantom{00-1} 3^x = 1 \\ 3^x = 3^1 \phantom{0}& \phantom{or00-1} 3^x = 3^0 \\ x = 1 \phantom{00} & \phantom{or00-10} x = 0 \end{align}

(a)

\begin{align} 5^{2x} - 6(5^x) + 5 & = 0 \\ [ a^{mn} = (a^m)^n] \phantom{00000000} (5^x)^2 - 6(5^x) + 5 & = 0 \\ \\ \text{When } & u = 5^x, \\ u^2 - 6u + 5 & = 0 \\ (u - 5)(u - 1) & =0 \\ \\ u - 5 = 0 \phantom{00}&\text{or}\phantom{00} u - 1 = 0 \\ u = 5 \phantom{00} &\phantom{or00-1} u = 1 \\ \\ \text{Since } & u = 5^x, \\ 5^x = 5 \phantom{00}& \text{or} \phantom{00-1} 5^x = 1 \\ 5^x = 5^1 \phantom{(.} & \phantom{or00-1} 5^x = 5^0 \\ \\ x = 1 \phantom{00}& \phantom{or00-10} x = 0 \end{align}

(b)

\begin{align} 2^{2x} - 10(2^x) + 16 & = 0 \\ [ a^{mn} = (a^m)^n] \phantom{00000000} (2^x)^2 - 10(2^x) + 16 & = 0 \\ \\ \text{When } & u = 2^x, \\ u^2 - 10u + 16 & = 0 \\ (u - 8)(u - 2) & = 0 \\ \\ u - 8 = 0 \phantom{00}&\text{or}\phantom{00} u -2 = 0 \\ u = 8 \phantom{00}&\phantom{or00-2} u = 2 \\ \\ \text{Since } & u = 2^x, \\ 2^x = 8 \phantom{000}&\text{or}\phantom{00-2} 2^x = 2 \\ 2^x = 2^3 \phantom{0(.}& \phantom{or00-2} 2^x = 2^1 \\ \\ x = 3 \phantom{000} & \phantom{or00-2(.} x = 1 \end{align}

\begin{align} 2^{2x + 1} & = 3(2^x) + 2 \\ [ a^{m + n} = (a^m)(a^n)] \phantom{000000000} (2^{2x})(2^1) & = 3(2^x) + 2 \\ 2(2^{2x}) & = 3(2^x) + 2 \\ [ a^{mn} = (a^m)^n] \phantom{0000000000} 2(2^x)^2 & = 3(2^x) + 2 \\ \\ \text{When } & u = 2^x, \\ 2u^2 & = 3u + 2 \\ 2u^2 - 3u - 2 & = 0 \\ (2u + 1)(u - 2) & = 0 \end{align} \begin{align} 2u + 1 & = 0 \phantom{0}&\text{or}\phantom{0000} u - 2 & = 0 \\ 2u & = -1 & u & = 2 \\ u & = -{1 \over 2} \\ \\ \text{Since } & u = 2^x, \\ 2^x & = -{1 \over 2} \text{ (N.A.)} & \text{or}\phantom{0000} 2^x & = 2 \\ & & 2^x & = 2^1 \\ \\ & & x & = 1 \end{align} $$ \therefore \text{Equation is only satisfied by only one value of } x $$

(a)

\begin{align} 5^x(25^{2y}) & = 1 \\ 5^x[(5^2)^{2y}] & = 5^0 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 5^x(5^{4y}) & = 5^0 \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 5^{x + 4y} & = 5^0 \\ \\ \therefore x + 4y & = 0 \\ x & = -4y \phantom{000} \text{ --- (1)} \end{align} \begin{align} 3^{5x}(9^y) & = {1 \over 9} \\ 3^{5x}[(3^2)^y] & = {1 \over 3^2} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 3^{5x} (3^{2y}) & = {1 \over 3^2} \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 3^{5x + 2y} & = {1 \over 3^2} \\ 3^{5x + 2y} & = 3^{-2} \phantom{00000000} \left[ {1 \over a^m} = a^{-m} \right] \\ \\ \therefore 5x + 2y & = -2 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 5(-4y) + 2y & = -2 \\ -20y + 2y & = -2 \\ -18y & = -2 \\ y & = {-2 \over -18} \\ & = {1 \over 9} \\ \\ \text{Substitute } & y = {1 \over 9} \text{ into (1),} \\ x & = -4 \left(1 \over 9\right) \\ & = -{4 \over 9} \\ \\ \therefore x & = -{4 \over 9}, y = {1 \over 9} \end{align}

(b)

\begin{align} {3^x \over 9^y} & = 27 \\ {3^x \over (3^2)^y} & = 3^3 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {3^x \over 3^{2y}} & = 3^3 \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 3^{x - 2y} & = 3^3 \\ \\ \therefore x - 2y & = 3 \\ x & = 2y + 3 \phantom{000} \text{ --- (1)} \end{align} \begin{align} 4^{2x}(2^{6y}) & = {1 \over 4} \\ (2^2)^{2x}(2^{6y}) & = {1 \over 2^2} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^{4x} (2^{6y}) & = {1 \over 2^2} \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 2^{4x + 6y} & = {1 \over 2^2} \\ 2^{4x + 6y} & = 2^{-2} \phantom{00000000} \left[ {1 \over a^m} = a^{-m} \right] \\ \\ \therefore 4x + 6y & = - 2 \\ 2x + 3y & = -1 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 2(2y + 3) + 3y & = -1 \\ 4y + 6 + 3y & = -1 \\ 4y + 3y & = -1 - 6 \\ 7y & = -7 \\ y & = {-7 \over 7} \\ & = -1 \\ \\ \text{Substitute } y = -1 \text{ into (1),} \\ x & = 2(-1) + 3 \\ & = 1 \\ \\ \therefore x & = 1, y = -1 \end{align}

(i)

\begin{align} \sqrt[n]{2 \times 4^m} & = \sqrt[n]{2^1 \times (2^2)^m} \\ & = \sqrt[n] {2^1 \times 2^{2m} } \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = \sqrt[n] {2^{1 + 2m}} \phantom{0000000000} [ a^m \times a^n = a^{m + n} ] \\ & = (2^{1 + 2m})^{1 \over n} \\ & = 2^{ {1 + 2m \over n}} \text{ (Shown)} \end{align}

(ii)

\begin{align} \sqrt[n]{2 \times 4^m} & = 8 \\ 2^{2m + 1 \over n} & = 8 \\ 2^{2m + 1 \over n} & = 2^3 \\ \\ \therefore {2m + 1 \over n} & = 3 \\ 2m + 1 & = 3n \\ 2m & = 3n - 1 \\ m & = {3n - 1 \over 2} \phantom{0} \text{ --- (1)} \end{align} \begin{align} {27^m \over 9^{n + 1}} & = 81 \\ {(3^3)^m \over (3^2)^{n + 1}} & = 3^4 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {3^{3m} \over 3^{2(n + 1)}} & = 3^4 \\ \left[ {a^m \over a^n} = a^{m - n} \right] \phantom{00000000} 3^{3m - 2(n + 1)} & = 3^4 \\ \\ \therefore 3m - 2(n + 1) & = 4 \\ 3m - 2n - 2 & = 4 \\ 3m - 2n & = 4 + 2 \\ 3m - 2n & = 6 \phantom{000} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(1) into (2),} \\ 3\left(3n - 1 \over 2\right) - 2n & = 6 \\ {3(3n - 1) \over 2} - 2n & = 6 \\ 3(3n - 1) - 4n & = 12 \\ 9n - 3 - 4n & = 12 \\ 9n - 4n & = 12 + 3 \\ 5n & = 15 \\ n & = {15 \over 5} \\ & = 3 \\ \\ \text{Substitute } & n = 3 \text{ into (1),} \\ m & = {3(3) - 1 \over 2} \\ & = 4 \\ \\ \therefore m & = 4, n = 3 \end{align}

(a)

\begin{align} 2(16^x) & = 5(4^x) - 2 \\ 2[(4^2)^x] & = 5(4^x) - 2 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2(4^{2x}) & = 5(4^x) - 2 \\ 2(4^x)^2 & = 5(4^x) - 2 \\ \\ \text{When } & u = 4^x, \\ 2u^2 & = 5u - 2 \\ 2u^2 - 5u + 2 & = 0 \\ (2u - 1)(u - 2) & = 0 \end{align} \begin{align} 2u - 1 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 2 & = 0 \\ 2u & = 1 & u & = 2 \\ u & = {1 \over 2} \\ \\ \\ \text{Since } & u = 4^x, \\ 4^x & = {1 \over 2} & 4^x & = 2 \\ 2^{2x} & = 2^{-1} & 2^{2x} & = 2^1 \\ \\ 2x & = -1 & 2x & = 1 \\ x & = -{1 \over 2} & x & = {1 \over 2} \end{align}

(b)

\begin{align} 7^{x + 1} - 2 & = 2(7^x) + 3 \\ [ a^{m + n} = (a^m)(a^n) ] \phantom{00000000} (7^x)(7^1) - 2 & = 2(7^x) + 3 \\ 7(7^x) - 2 & = 2(7^x) + 3 \\ \\ \text{When } & u = 7^x, \\ 7u - 2 & = 2u + 3 \\ 7u - 2u & = 3 + 2 \\ 5u & = 5 \\ u & = {5 \over 5} \\ & = 1 \\ \\ \text{Since } & u = 7^x, \\ 7^x & = 1 \\ 7^x & = 7^0 \\ \\ \therefore x & = 0 \end{align}

(c)

\begin{align} 9^{x + 1} + 1 & = 10(3^x) \\ [a^{m + n} = (a^m)(a^n)] \phantom{00000000} (9^x)(9^1) + 1 & = 10(3^x) \\ 9(9^x) + 1 & = 10(3^x) \\ 9(3^2)^x + 1 & = 10(3^x) \\ [ (a^m)^n = a^{mn} ] \phantom{000000000} 9(3^{2x}) + 1 & = 10(3^x) \\ 9(3^x)^2 + 1 & = 10(3^x) \\ \\ \text{When } & u = 3^x, \\ 9u^2 + 1 & = 10u \\ 9u^2 - 10u + 1 & = 0 \\ (9u - 1)(u - 1) & = 0 \end{align} \begin{align} 9u - 1 & = 0 \phantom{00}&\text{or}\phantom{00} u - 1 & = 0 \\ 9u & = 1 & u & = 1 \\ u & = {1 \over 9} \\ \\ \text{Since } & u = 3^x, \\ 3^x & = {1 \over 9} & 3^x & = 1 \\ 3^x & = {1 \over 3^2} & 3^x & = 3^0 \\ 3^x & = 3^{-2} \\ \\ \therefore x & = -2 & \therefore x & = 0 \end{align}

(d)

\begin{align} 6(3^{x - 1}) & = 3^4 - 3^x \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{00000000} 6\left( 3^x \over 3^1 \right) & = 3^4 - 3^x \\ 6 \left(3^x \over 3\right) & = 3^4 - 3^x \\ 2(3^x) & = 3^4 - 3^x \\ \\ \text{When } & u = 3^x, \\ 2u & = 3^4 - u \\ 2u + u & = 3^4 \\ 3u & = 81 \\ u & = {81 \over 3} \\ & = 27 \\ \\ \text{Since } & u = 3^x, \\ 3^x & = 27 \\ 3^x & = 3^3 \\ \\ \therefore x & = 3 \end{align}

(e)

\begin{align} (\sqrt{9})^{2x} - 3^{x + 2} & = 3^x - 9 \\ 3^{2x} - 3^{x + 2} & = 3^x - 9 \\ [ a^{mn} = (a^m)^n] \phantom{00000000} (3^x)^2 - 3^{x + 2} & = 3^x - 9 \\ [ a^{m + n} = (a^m)(a^n) ] \phantom{00000000} (3^x)^2 - (3^x)(3^2) & = 3^x - 9 \\ (3^x)^2 - (3^x)(9) & = 3^x - 9 \\ (3^x)^2 - 9(3^x) & = 3^x - 9 \\ \\ \text{When } & u = 3^x, \\ u^2 - 9u & = u - 9 \\ u^2 - 9u - u + 9 & = 0 \\ u^2 - 10u + 9 & = 0 \\ (u - 9)(u - 1) & = 0 \end{align} \begin{align} u - 9 & = 0 \phantom{00}&\text{or}\phantom{00} u - 1 & = 0 \\ u & = 9 & u & = 1 \\ \\ \text{Since } & u = 3^x, \\ 3^x & = 9 & 3^x & = 1 \\ 3^x & = 3^2 & 3^x & = 3^0 \\ \\ \therefore x & =2 & \therefore x & = 0 \end{align}

(i)

\begin{align} \sqrt{8^x}(4^y) & = 32\sqrt{2} \\ (8^x)^{1 \over 2} (4^y) & = 32(2)^{1 \over 2} \\ [(2^3)^x]^{1 \over 2} [(2^2)^y] & = (2^5)(2)^{1 \over 2} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} (2^{3x})^{1 \over 2} (2^{2y}) & = (2^5)(2^{1 \over 2}) \\ (2^{3x \over 2})(2^{2y}) & = (2^5)(2^{1 \over 2}) \\ 2^{{3x \over 2} + 2y} & = 2^{5 + {1 \over 2}} \phantom{00000000} [ (a^m)(a^n) = a^{m + n}] \\ \\ \therefore {3x \over 2} + 2y & = 5 + {1 \over 2} \\ {3x \over 2} + 2y & = {11 \over 2} \\ 3x + 4y & = 11 \end{align}

(ii)

\begin{align} {3^y \over 9^{x^2}} & = {1 \over 3}(3^x) \\ {3^y \over (3^2)^{x^2}} & = {3^x \over 3} \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} {3^y \over 3^{2x^2}} & = {3^x \over 3^1} \\ 3^{y - 2x^2} & = 3^{x - 1} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ \\ \therefore y - 2x^2 & = x - 1 \\ y & = 2x^2 + x - 1 \end{align}

(iii)

\begin{align} 3x + 4y & = 11 \phantom{0} \text{ --- (1)} \\ \\ y & = 2x^2 + x - 1 \phantom{0} \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3x + 4(2x^2 + x - 1) & = 11 \\ 3x + 8x^2 + 4x - 4 & = 11 \\ 8x^2 + 7x - 15 & = 0 \\ (8x + 15)(x - 1) & = 0 \end{align} \begin{align} 8x + 15 & = 0 \phantom{00}&\text{or}\phantom{0000} x - 1 & = 0 \\ 8x & = -15 & x & =1 \\ x & = -{15 \over 8} \\ & = -1{7 \over 8} \\ \\ \text{Substitute } & \text{into (2),} \\ y & = 2\left(-{15 \over 8}\right)^2 + \left(-{15 \over 8}\right) + 1 & y & = 2(1)^2 + (1) - 1 \\ & = 4{5 \over 32} & & = 2 \end{align}

\begin{align} {r^2 \over 4}(3x)^r \left( 2 \over 9x^2 \right)^{6 - r} & = {k \over x^3} \\ {r^2 \over 2^2}(3x)^r \left( 2 \over 3^2 x^2 \right)^{6 - r} & = {k \over x^3} \\ {r^2 \over 2^2}(3^r)(x^r)\left[(2)(3^{-2})(x^{-2})\right]^{6 -r} & = {k \over x^3} \\ (r^2)(2^{-2})(3^r)(x^r)(2^{6 - r})(3^{-2(6 - r)})(x^{-2(6 - r)}) & = {k \over x^3} \\ (r^2)(2^{-2})(3^r)(x^r)(2^{6 - r})(3^{-12 + 2r})(x^{-12 + 2r}) & = (k)(x^{-3}) \\ (r^2)(2^{-2})(2^{6 - r})(3^r)(3^{-12 + 2r})(x^r)(x^{-12 + 2r}) & = (k)(x^{-3}) \\ (r^2)(2^{-2 + 6 - r})(3^{r - 12 + 2r})(x^{r - 12 + 2r}) & = (k)(x^{-3}) \\ (r^2)(2^{4 - r})(3^{3r - 12})(x^{3r - 12}) & = (k)(x^{-3}) \\ \\ \text{Comparing the powers } & \text{of } x, \\ x^{3r - 12} & = x^{-3} \\ \\ \therefore 3r - 12 & = -3 \\ 3r & = 9 \\ r & = {9 \over 3} \\ & = 3 \\ \\ \text{When } & r = 3, \\ ((3)^2)(2^{4 - 3})(3^{3(3) - 12})(x^{3(3) - 12}) & = (k)(x^{-3}) \\ (9)(2)(3^{-3})(x^{-3}) & = (k)(x^{-3}) \\ 18\left(1 \over 27\right)(x^{-3}) & = (k)(x^{-3}) \\ {18 \over 27}x^{-3} & = kx^{-3} \\ {2 \over 3} & = k \\ \\ \therefore r & = 3, k = {2 \over 3} \end{align}

\begin{align} 64(4^y) & = 16^x \\ (4^3)(4^y) & = (4^2)^x \\ [ (a^m)(a^n) = a^{m + n} ] \phantom{00000000} 4^{3 + y} & = (4^2)^x \\ 4^{3 + y} & = 4^{2x} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ \\ \therefore 3 + y & = 2x \\ y & = 2x - 3 \phantom{000} \text{ --- (1)} \end{align} \begin{align} 3^y & = 4(3^{x - 2}) - 1 \\ 3^y & = 4\left( 3^x \over 3^2 \right) - 1 \phantom{00000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ 3^y & = 4\left( 3^x \over 9 \right) - 1 \\ 3^y & = {4 \over 9}(3^x) - 1 \phantom{0} \text{ --- (2)} \end{align}
\begin{align} \text{Substitute } & \text{(1) into (2),} \\ 3^{2x - 3} & = {4 \over 9}(3^x) - 1 \\ {3^{2x} \over 3^3} & = {4 \over 9}(3^x) - 1 \\ {3^{2x} \over 27} & = {4 \over 9}(3^x) - 1 \\ {(3^x)^2 \over 27} & = {4 \over 9}(3^x) - 1 \\ \\ \text{When } & u = 3^x, \\ {u^2 \over 27} & = {4 \over 9}u - 1 \\ u^2 & = 12u - 27 \\ u^2 - 12u + 27 & = 0 \\ (u - 3)(u - 9) & = 0 \end{align} \begin{align} u - 3 & = 0 \phantom{00}&\text{or}\phantom{0000} u - 9 & = 0 \\ u & = 3 & u & = 9 \\ \\ \text{Since } & u = 3^x, \\ 3^x & = 3 & 3^x & = 9 \\ 3^x & = 3^1 & 3^x & = 3^2 \\ \\ x & = 1 & x & = 2 \\ \\ \text{Substitute } & \text{into (1),} \\ \\ y & = 2(1) - 3 & y & = 2(2) - 3 \\ & = -1 & & = 1 \end{align}

\begin{align} x^{3 \over 2} - 8x^{-{3 \over 2}} & = 7 \\ x^{3 \over 2} - 8\left( 1 \over x^{3 \over 2} \right) & = 7 \phantom{00000000} \left[ a^{-m} = {1 \over a^m} \right] \\ \\ x^{3 \over 2} - {8 \over x^{3 \over 2}} & = 7 \\ \\ \text{When } & u = x^{3 \over 2}, \\ u - {8 \over u} & = 7 \\ u^2 - 8 & = 7u \\ u^2 - 7u - 8 & = 0 \\ (u - 8)(u + 1) & = 0 \end{align} \begin{align} u - 8 & = 0 \phantom{00}&\text{or}\phantom{0000} u + 1 & = 0 \\ u & = 8 & u & = -1 \\ \\ \text{Since } & u = x^{3 \over 2}, \\ x^{3 \over 2} & = 8 & x^{3 \over 2} & = -1 \text{ (N.A)} \\ (x^{3 \over 2})^2 & = 8^2 \\ x^3 & = 64 \\ x & = \sqrt[3]{64} \\ & = 4 \end{align}

\begin{align} x = 2, y = 5 & \text{ only if } a \ne b \\ \\ \text{If } & a = b = 2, \\ (2)^x (2)^y & = (2)^2 (2)^5 \\ 2^{x + y} & = 2^7 \\ \\ \text{Then }x = 2, y = 5 & \text{ is just one possible solution} \end{align}