(a)

\begin{align} (2^3 \times 3^2)^0 & = (2^3)^0 \times (3^2)^0 \phantom{00000} [(a \times b)^m = a^m \times b^m] \\ & = 2^{3(0)} \times 3^{2(0)} \phantom{000000} [(a^m)^n = a^{mn}] \\ & = 2^0 \times 3^0 \\ & = 1 \times 1 \\ & = 1 \end{align}

(b)

\begin{align} (2\sqrt{3})^4 & = [2(3^{1 \over 2})]^4 \\ & = (2^4) [(3^{1 \over 2})^4] \phantom{00000} [(a \times b)^m = a^m \times b^m] \\ & = (2^4) [3^{{1 \over 2}(4)}] \phantom{00000.} [(a^m)^n = a^{mn}] \\ & = (2^4) (3^2) \\ & = (16)(9) \\ & = 144 \end{align}

(c)

\begin{align} {9^{1 \over 3} \times 3^{1 \over 3} \over 6} & = { (9 \times 3)^{1 \over 3} \over 6} \phantom{00000} [a^m \times b^m = (a \times b)^m] \\ & = {27^{1 \over 3} \over 6} \\ & = {\sqrt[3]{27} \over 6} \\ & = {3 \over 6} \\ & = {1 \over 2} \end{align}

(d)

\begin{align} 25^{1 \over 4} \times 5^{1 \over 3} \times 5^{1 \over 6} & = (5^2)^{1 \over 4} \times 5^{1 \over 3} \times 5^{1 \over 6} \\ & = 5^{2({1 \over 4})} \times 5^{1 \over 3} \times 5^{1 \over 6} \phantom{00000} [ (a^m)^n = a^{mn}] \\ & = 5^{1 \over 2} \times 5^{1 \over 3} \times 5^{1 \over 6} \\ \\ & = 5^{{1 \over 2} + {1 \over 3} + {1 \over 6}} \phantom{00000000000.} [ a^m \times a^n = a^{m + n} ] \\ & = 5^{1} \\ & = 5 \end{align}

(e)

\begin{align} {3^{1 \over 3} \times 3^0 \times 9^{1 \over 3} \over 27^{2 \over 3}} & = {3^{1 \over 3} \times 1 \times 9^{1 \over 3} \over 27^{2 \over 3}} \\ & = {3^{1 \over 3} \times 9^{1 \over 3} \over 27^{2 \over 3}} \\ & = {3^{1 \over 3} \times (3^2)^{1 \over 3} \over 27^{2 \over 3}} \\ & = {3^{1 \over 3} \times 3^{2 \over 3} \over 27^{2 \over 3}} \phantom{00000} [(a^m)^n = a^{mn}] \\ & = {3^{{1 \over 3} + {2 \over 3}} \over 27^{2 \over 3}} \phantom{0000000.} [ a^m \times a^n = a^{m + n}] \\ & = {3^1 \over 27^{2 \over 3}} \\ & = {3 \over (\sqrt[3]{27})^2} \phantom{0000000} [ a^{m \over n} = (\sqrt[n]{a})^m] \\ & = {3 \over 9} \\ & = {1 \over 3} \end{align}

(a)

\begin{align} \text{When } & x = 9, \\ y & = 2(9)^{3 \over 2} \\ & = 2(27) \\ & = 54 \end{align}

(b)

\begin{align} \text{When } & y = 16, \\ 16 & = 2x^{3 \over 2} \\ {16 \over 2} & = x^{3 \over 2} \\ 8 & = x^{3 \over 2} \\ 8^2 & = (x^{3 \over 2})^2 \\ 64 & = x^{{3 \over 2}(2)} \phantom{00000} [ (a^m)^n = a^{mn} ] \\ 64 & = x^3 \\ \sqrt[3]{64} & = x \\ 4 & = x \end{align}

(a)

\begin{align} 16^x & = (2^4)^x \\ & = 2^{4x} \phantom{00000} [ (a^m)^n = a^{mn}] \\ & = (2^x)^4 \\ & = (u)^4 \\ & = u^4 \end{align}

(b)

\begin{align} 2^{2x + 3} & = 2^{2x} \times 2^3 \phantom{000000} [ a^{m + n} = a^m + a^n] \\ & = (2^x)^2 \times 2^3 \phantom{0000.} [ a^{mn} = (a^m)^n ] \\ & = (u)^2 \times 8 \\ & = u^2 \times 8 \\ & = 8u^2 \end{align}

(c)

\begin{align} 4^{x - 1} & = {4^x \over 4^1} \phantom{00000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ & = {(2^2)^x \over 4} \\ & = {2^{2x} \over 4} \phantom{00000} [ (a^m)^n = a^{mn} ] \\ & = {(2^x)^2 \over 4} \\ & = {(u)^2 \over 4} \\ & = {u^2 \over 4} \end{align}

(d)

\begin{align} 4^{ {1 \over 2} - x} & = {4^{1 \over 2} \over 4^x} \phantom{0000000} \left[ a^{m - n} = {a^m \over a^n} \right] \\ & = { \sqrt{4} \over (2^2)^x} \\ & = {2 \over 2^{2x}} \phantom{00000000} [ (a^m)^n = a^{mn} ] \\ & = {2 \over (2^x)^2 } \\ & = {2 \over (u)^2} \\ & = {2 \over u^2} \end{align}

(a)

\begin{align} 8 \times 2^x \div \sqrt{2^y} & = 2^3 \times 2^x \div \sqrt{2^y} \\ & = 2^{3 + x} \div \sqrt{2^y} \phantom{0000000} [a^{m} \times a^n = a^{m + n}] \\ & = 2^{3 + x} \div (2^y)^{1 \over 2} \phantom{00000(} [\sqrt[n]{a} = a^{1 \over n}] \\ & = 2^{3 + x} \div 2^{{1 \over 2}y} \phantom{0000000.} [(a^m)^n = a^{mn}] \\ & = 2^{3 + x - {1 \over 2}y} \phantom{00000000(0} [a^m \div a^n = a^{m - n} ] \end{align}

(b)

\begin{align} 2^x \times 4^{2x + 1} \times {1 \over 2}(16^x) & = 2^x \times (2^2)^{2x + 1} \times {1 \over 2}(16^x) \\ & = 2^x \times 2^{2(2x + 1)} \times {1 \over 2}(16^x) \phantom{0000000} [(a^m)^n = a^{mn}] \\ & = 2^x \times 2^{2(2x + 1)} \times (2^{-1})(2^4)^x \\ & = 2^x \times 2^{2(2x + 1)} \times (2^{-1})(2^{4x}) \\ & = 2^x \times 2^{2(2x + 1)} \times 2^{-1 + 4x} \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ & = 2^{x + 2(2x + 1) + (-1 + 4x)} \phantom{000000000000} [ a^{m} \times a^n = a^{m + n} ] \\ & = 2^{x + 4x + 2 - 1 + 4x} \\ & = 2^{9x + 1} \end{align}

(c)

\begin{align} {1 \over 4^{x - 1}} \times \sqrt{8^y \over 32} & = {1 \over (2^2)^{x - 1}} \times \sqrt{(2^3)^y \over 2^5} \phantom{000000} [ (a^m)^n = a^{mn} ] \\ & = {1 \over 2^{2(x - 1)}} \times \sqrt{2^{3y} \over 2^5} \\ & = {1 \over 2^{2x - 2}} \times \sqrt{2^{3y - 5}} \phantom{00000000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = (2^{2x - 2})^{-1} \times \sqrt{2^{3y - 5}} \phantom{00000.} \left[ {1 \over a^m} = a^{-m} \right] \\ & = 2^{-(2x - 2)} \times (2^{3y - 5})^{1 \over 2} \\ & = 2^{-2x + 2} \times 2^{{1 \over 2}(3y - 5)} \\ & = 2^{-2x + 2} \times 2^{{3 \over 2}y - {5 \over 2}} \\ & = 2^{-2x + 2 + {3 \over 2}y - {5 \over 2}} \phantom{000000000000} [a^m \times a^n = a^{m + n}] \\ & = 2^{-2x + {3 \over 2}y - {1 \over 2}} \end{align}

(a)

\begin{align} {8^{x + 2} - 34(2^{3x}) \over (\sqrt{8})^{2x + 1} } & = { (2^3)^{x + 2} - 34(2^{3x}) \over (\sqrt{2^3})^{2x + 1} } \\ & = { 2^{3(x + 2)} - 34(2^{3x}) \over (2^{3 \over 2})^{2x + 1} } \phantom{000000} [\sqrt[n]{a^m} = a^{m \over n}] \\ & = { 2^{3x + 6} - 34(2^{3x}) \over 2^{{3 \over 2}(2x + 1)} } \\ & = { 2^{3x} (2^6) - 34(2^{3x}) \over 2^{3x + {3 \over 2}} } \\ & = { 64(2^{3x}) - 34(2^{3x}) \over 2^{3x + {3 \over 2}} } \\ & = { 30(2^{3x}) \over 2^{3x + {3 \over 2}} } \\ & = 30 \left( 2^{3x} \over 2^{3x + {3 \over 2}} \right) \\ & = 30 (2^{3x} \div 2^{3x + {3 \over 2}} ) \\ & = 30 (2^{3x - (3x + {3 \over 2})}) \\ & = 30 (2^{3x - 3x - {3 \over 2}}) \\ & = 30 (2^{-{3 \over 2}}) \\ & = 15(2)(2^{-{3 \over 2}}) \\ & = 15(2^1 \times 2^{-{3 \over 2}}) \\ & = 15 \left[ 2^{1 + \left( - {3 \over 2} \right) } \right] \\ & = 15 (2^{-{1 \over 2}}) \\ \\ & = 15\left( 1 \over 2^{1 \over 2} \right) \phantom{000000000} \left[ a^{-m} = {1 \over a^m} \right] \\ & = 15 \left( 1 \over \sqrt{2} \right) \\ & = {15 \over \sqrt{2}} \\ & = {15 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} \phantom{000000000} [\text{Rationalise denominator}] \\ & = {15\sqrt{2} \over 2} \\ & = {15 \over 2}\sqrt{2} \end{align}

(b)

\begin{align} {5^{3x} \times 25^{1 - x} \over 5^{2x + 1} + 3(25^x) + 17(5^{2x}) } & = {5^{3x} \times (5^2)^{1 - x} \over (5^{2x})(5) + 3[(5^2)^x] + 17(5^{2x}) } \\ & = {5^{3x} \times 5^{2(1 - x)} \over 5(5^{2x}) + 3(5^{2x}) + 17(5^{2x}) } \\ & = {5^{3x} \times 5^{2 - 2x} \over (5^{2x})(5 + 3 + 17) } \\ & = {5^{3x + 2 - 2x} \over (5^{2x})(25) } \\ & = {5^{x + 2} \over (5^{2x})(5^2) } \\ & = {5^{x + 2} \over 5^{2x + 2} } \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ & = 5^{x + 2 - (2x + 2)} \phantom{0000} \left[ {a^m \over a^n} = a^{m - n} \right] \\ & = 5^{x + 2 - 2x - 2} \\ & = 5^{-x} \end{align}

(a)

\begin{align} (\sqrt{a})^4 \times (\sqrt[4]{a})^{12} \div a^4 & = (a^{1 \over 2})^4 \times (a^{1 \over 4})^{12} \div a^4 \\ & = a^{{1 \over 2}(4)} \times a^{{1 \over 4}(12)} \div a^4 \\ & = a^2 \times a^3 \div a^4 \\ & = a^{2 + 3} \div a^4 \\ & = a^5 \div a^4 \\ & = a^{5 - 4} \\ & = a^1 \\ & = a \end{align}

(b)

\begin{align} {2(\sqrt[3]{a}) + (\sqrt[6]{a})^2 \over \sqrt[9]{a^2}} & = {2(a^{1 \over 3}) + (a^{1 \over 6})^2 \over (a^2)^{1 \over 9}} \\ & = {2(a^{1 \over 3}) + a^{{1 \over 6}(2)} \over a^{2({1 \over 9})} } \\ & = {2a^{1 \over 3} + a^{1 \over 3} \over a^{2 \over 9}} \\ & = {3a^{1 \over 3} \over a^{2 \over 9}} \\ & = 3\left( a^{1 \over 3} \over a^{2 \over 9} \right) \\ & = 3 ( a^{{1 \over 3} - {2 \over 9}} ) \\ & = 3 ( a^{1 \over 9}) \\ & = 3(\sqrt[9]{a}) \end{align}

(a)

\begin{align} & (a^{1 \over 3} + b^{1 \over 3})(a^{2 \over 3} - a^{1 \over 3}b^{1 \over 3} + b^{2 \over 3}) \\ = \phantom{0} & (a^{1 \over 3})(a^{2 \over 3}) - (a^{1 \over 3})(a^{1 \over 3}b^{1 \over 3}) + (a^{1 \over 3})(b^{2 \over 3}) + (a^{2 \over 3})(b^{1 \over 3}) - (b^{1 \over 3})(a^{1 \over 3}b^{1 \over 3}) + (b^{1 \over 3})(b^{2 \over 3}) \\ = \phantom{0} & a^{{1 \over 3} + {2 \over 3}} - a^{{1 \over 3} + {1 \over 3}}b^{1 \over 3} + a^{1 \over 3}b^{2 \over 3} + a^{2 \over 3}b^{1 \over 3} - a^{1 \over 3}b^{{1 \over 3} + {1 \over 3}} + b^{{1 \over 3} + {2 \over 3}} \\ = \phantom{0} & a - a^{2 \over 3}b^{1 \over 3} + a^{1 \over 3}b^{2 \over 3} + a^{2 \over 3}b^{1 \over 3} - a^{1 \over 3}b^{2 \over 3} + b \\ = \phantom{0} & a + b \end{align}

(b) Use the identity (x + y)(x - y) = x² - y²

\begin{align} (\sqrt[3]{\sqrt{a^3 + b^2} + b})(\sqrt[3]{\sqrt{a^3 + b^2} - b}) & = (\sqrt{a^3 + b^2} + b)^{1 \over 3} (\sqrt{a^3 + b^2} - b)^{1 \over 3} \\ & = [(\underbrace{\sqrt{a^3 + b^2}}_{x} + \underbrace{b}_{y})(\underbrace{\sqrt{a^3 + b^2}}_{x} - \underbrace{b}_{y})]^{1 \over 3} \\ & = \left[ \left(\sqrt{a^3 + b^2} \right)^2 - (b)^2 \right]^{1 \over 3} \\ & = (a^3 + b^2 - b^2)^{1 \over 3} \\ & = (a^3)^{1 \over 3} \\ & = a^{3({1 \over 3})} \\ & = a^1 \\ & = a \end{align}

(c)

\begin{align} (\sqrt[3]{3} - \sqrt[3]{2})(\sqrt[3]{9} + \sqrt[3]{6} + \sqrt[3]{4}) & = (3^{1 \over 3} - 2^{1 \over 3})(9^{1 \over 3} + 6^{1 \over 3} + 4^{1 \over 3}) \\ & = (3^{1 \over 3})(9^{1 \over 3}) + (3^{1 \over 3})(6^{1 \over 3}) + (3^{1 \over 3})(4^{1 \over 3}) - (2^{1 \over 3})(9^{1 \over 3}) - (2^{1 \over 3})(6^{1 \over 3}) - (2^{1 \over 3})(4^{1 \over 3}) \\ & = 27^{1 \over 3} + 18^{1 \over 3} + 12^{1 \over 3} - 18^{1 \over 3} - 12^{1 \over 3} - 8^{1 \over 3} \phantom{000000} [a^m \times b^m = (a \times b)^m ] \\ & = 27^{1 \over 3} - 8^{1 \over 3} \\ & = \sqrt[3]{27} - \sqrt[3]{8} \\ & = 3 - 2 \\ & = 1 \end{align}

(d) Use the identity (x + y)(x - y) = x² - y²

\begin{align} (a^{1 \over 2} + \sqrt{2}a^{1 \over 4}b^{1 \over 4} + b^{1 \over 2})(a^{1 \over 2} - \sqrt{2}a^{1 \over 4}b^{1 \over 4} + b^{1 \over 2}) & = \left[\underbrace{(a^{1 \over 2} + b^{1 \over 2})}_{x} + \underbrace{\sqrt{2}a^{1 \over 4}b^{1 \over 4}}_{y} \right] \left[\underbrace{(a^{1 \over 2} + b^{1 \over 2})}_{x} - \underbrace{\sqrt{2}a^{1 \over 4}b^{1 \over 4}}_{y} \right] \\ & = (a^{1 \over 2} + b^{1 \over 2})^2 - (\sqrt{2}a^{1 \over 4}b^{1 \over 4})^2 \\ & = (a^{1 \over 2})^2 + 2(a^{1 \over 2})(b^{1 \over 2}) + (b^{1 \over 2})^2 - (2^{1 \over 2}a^{1 \over 4}b^{1 \over 4})^2 \\ & = a^{{1 \over 2}(2)} + 2a^{1 \over 2}b^{1 \over 2} + b^{{1 \over 2}(2)} - 2^{{1 \over 2}(2)} a^{{1 \over 4}(2)} b^{{1 \over 4}(2)} \\ & = a + 2a^{1 \over 2}b^{1 \over 2} + b - 2a^{1 \over 2}b^{1 \over 2} \\ & = a + b \end{align}

\begin{align} (\sqrt{3})^7 + (\sqrt{3})^5 + (\sqrt{3})^3 + 42(\sqrt{3}) & = 3^k \\ (\sqrt{3})^{6 + 1} + (\sqrt{3})^{4 + 1} + (\sqrt{3})^{2 + 1} + 42(\sqrt{3}) & = 3^k \\ (\sqrt{3})^6(\sqrt{3}) + (\sqrt{3})^4(\sqrt{3}) + (\sqrt{3})^2(\sqrt{3}) + 42(\sqrt{3}) & = 3^k \\ 27\sqrt{3} + 9\sqrt{3} + 3\sqrt{3} + 42\sqrt{3} & = 3^k \\ \sqrt{3} (27 + 9 + 3 + 42) & = 3^k \\ \sqrt{3} (81) & = 3^k \\ \sqrt{3} (3^4) & = 3^k \\ (3^{1 \over 2})(3^4) & = 3^k \\ (3^{0.5})(3^4) & = 3^k \\ 3^{0.5 + 4} & = 3^k \\ 3^{4.5} & = 3^k \\ \\ \therefore k & = 4.5 \end{align}

(a)

\begin{align} (36p^6)^{2 \over 3} \times \sqrt{54p^5} & = (36p^6)^{2 \over 3} \times (54p^5)^{1 \over 2} \\ & = (4 \times 9 \times p^6)^{2 \over 3} \times (2 \times 27 \times p^5)^{1 \over 2} \\ & = (2^2 \times 3^2 \times p^6)^{2 \over 3} \times (2 \times 3^3 \times p^5)^{1 \over 2} \\ & = (2^2)^{2 \over 3} \times (3^2)^{2 \over 3} \times (p^6)^{2 \over 3} \times (2)^{1 \over 2} \times (3^3)^{1 \over 2} \times (p^5)^{1 \over 2} \\ & = 2^{4 \over 3} \times 3^{4 \over 3} \times p^4 \times 2^{1 \over 2} \times 3^{3 \over 2} \times p^{5 \over 2} \\ & = 2^{4 \over 3} \times 2^{1 \over 2} \times 3^{4 \over 3} \times 3^{3 \over 2} \times p^4 \times p^{5 \over 2} \\ & = 2^{{4 \over 3} + {1 \over 2}} 3^{{4 \over 3} + {3 \over 2}} p^{4 + {5 \over 2}} \\ & = 2^{11 \over 6}3^{17 \over 6} p^{13 \over 2} \end{align} \begin{align} 2^{11 \over 6}3^{17 \over 6} p^{13 \over 2} & = 2^x 3^y p^z \\ \\ \therefore x & = {11 \over 6} = 1{5 \over 6} \\ \\ \therefore y & = {17 \over 6} = 2{5 \over 6} \\ \\ \therefore z & = {13 \over 2} = 6{1 \over 2} \end{align}

(b)

\begin{align} (144p^2)^{5 \over 2} \div (24p^{-2})^{3 \over 2} & = {(144p^2)^{5 \over 2} \over (24p^{-2})^{3 \over 2}} \\ & = {(16 \times 9 \times p^2)^{5 \over 2} \over (8 \times 3 \times p^{-2})^{3 \over 2}} \\ & = {(2^4 \times 3^2 \times p^2)^{5 \over 2} \over (2^3 \times 3^1 \times p^{-2})^{3 \over 2}} \\ & = {(2^4)^{5 \over 2} \times (3^2)^{5 \over 2} \times (p^2)^{5 \over 2} \over (2^3)^{3 \over 2} \times (3^1)^{3 \over 2} \times (p^{-2})^{3 \over 2}} \\ & = {2^{10} \times 3^5 \times p^5 \over 2^{9 \over 2} \times 3^{3 \over 2} \times p^{-3}} \\ & = 2^{10 - {9 \over 2}} 3^{5 - {3 \over 2}} p^{5 - (-3)} \\ & = 2^{11 \over 2} 3^{7 \over 2} p^8 \end{align} \begin{align} 2^{11 \over 2} 3^{7 \over 2} p^8 & = 2^x 3^y p^z \\ \\ \therefore x & = {11 \over 2} = 5{1 \over 2} \\ \\ \therefore y & = {7 \over 2} = 3{1 \over 2} \\ \\ \therefore z & = 8 \end{align}

(i)

\begin{align} 2^{n + 1} - 3(2^n) + 4(2^{n + 2}) & = (2^n)(2^1) - 3(2^n) + 4[(2^n)(2^2)] \phantom{00000000} [a^{m + n} = (a^m)(a^n)] \\ & = 2(2^n) - 3(2^n) + 4[(2^n)(4)] \\ & = 2(2^n) - 3(2^n) + 16(2^n) \\ & = (2^n)(2 - 3 + 16) \\ & = (2^n)(15) \\ & = 15(2^n) \end{align}

(ii)

\begin{align} {2^{n + 1} - 3(2^n) + 4(2^{n + 2}) \over 15} & = {15(2^n) \over 15} \\ \\ & = 2^n \\ \\ \therefore \text{Expression is divisble by } & 15 \text{ for } n > 0 \end{align}

\begin{align} 4^n + 5(4^{n + 1}) - 2^{2n + 1} & = (2^2)^n + 5[(2^2)^{n + 1}] - 2^{2n + 1} \\ & = 2^{2n} + 5(2^{2(n + 1)}) - 2^{2n + 1} \phantom{0000000000} [ (a^m)^n = a^{mn} ] \\ & = 2^{2n} + 5(2^{2n + 2}) - 2^{2n + 1} \\ & = 2^{2n} + 5(2^{2n})(2^2) - (2^{2n})(2^1) \phantom{0000000} [ a^{m + n} = (a^m)(a^n)] \\ & = 2^{2n} + 5(2^{2n})(4) - (2^{2n})(2) \\ & = 2^{2n} + 20(2^{2n}) - 2(2^{2n}) \\ & = (2^{2n})(1 + 20 - 2) \\ & = (2^{2n})(19) \\ & = 19(2^{2n}) \end{align} \begin{align} {4^n + 5(4^{n + 1}) - 2^{2n + 1} \over 19} & = {19(2^{2n}) \over 19} \\ \\ & = 2^{2n} \\ \\ \therefore \text{Expression is divisible by } & 19 \text{ for } n > 0 \end{align}

\begin{align} P & = (2 + \sqrt{5})^{2n - 1} \\ & = (2 + \sqrt{5})^{2n + (-1)} \\ & = (2 + \sqrt{5})^{2n} \times (2 + \sqrt{5})^{-1} \phantom{000000000} [a^{m + n} = a^m \times a^n] \\ & = (2 + \sqrt{5})^{2n} \times {1 \over 2 + \sqrt{5}} \\ & = {(2 + \sqrt{5})^{2n} \over 2 + \sqrt{5}} \\ & = {[(2 + \sqrt{5})^2]^n \over 2 + \sqrt{5}} \phantom{0000000000000000000} [ a^{mn} = (a^m)^n] \\ & = {[(2)^2 + 2(2)(\sqrt{5}) + (\sqrt{5})^2]^n \over 2 + \sqrt{5}} \\ & = {(4 + 4\sqrt{5} + 5)^n \over 2 + \sqrt{5}} \\ & = {(9 + 4\sqrt{5})^n \over 2 + \sqrt{5}} \\ & = {(9 + \sqrt{16}\sqrt{5})^n \over 2 + \sqrt{5}} \\ & = {(9 + \sqrt{80})^n \over 2 + \sqrt{5}} \\ & = {(9 + \sqrt{80})^n \over \sqrt{5} + 2} \times {\sqrt{5} - 2 \over \sqrt{5} - 2} \phantom{00000000000} [\text{Rationalise denominator}] \\ & = {(9 + \sqrt{80})^n(\sqrt{5} - 2) \over (\sqrt{5} + 2)(\sqrt{5} - 2)} \\ & = {(\sqrt{5} - 2)(9 + \sqrt{80})^n \over (\sqrt{5})^2 - (2)^2} \\ & = {(\sqrt{5} - 2)(9 + \sqrt{80})^n \over 5 - 4} \\ & = {(\sqrt{5} - 2)(9 + \sqrt{80})^n \over 1} \\ \\ & = (\sqrt{5} - 2)(9 + \sqrt{80})^n \phantom{0} \text{ (Shown)} \end{align}

\begin{align} (8x^2)^{8 - r} \left( 1 \over 2x \right)^r & = (8)^{8 - r} (x^2)^{8 - r} [(2x)^{-1}]^r \\ & = (2^3)^{8 - r} (x^2)^{8 - r} (2x)^{-1(r)} \phantom{000000000.} [ (a^m)^n = a^{mn}] \\ & = (2^{3(8 - r)}) (x^{2(8 - r)}) (2x)^{-r} \\ & = (2^{24 - 3r}) (x^{16 - 2r}) (2^{-r}) (x^{-r}) \phantom{0000000} [ (ab)^m = a^m \times b^m] \\ & = (2^{24 - 3r})(2^{-r}) (x^{16 - 2r})(x^{-r}) \\ & = (2^{24 - 3r + (-r)}) (x^{16 - 2r + (-r)}) \phantom{00000000} [ (a^m)(a^n) = a^{m + n} ] \\ & = (2^{24 - 3r - r}) (x^{16 - 2r - r}) \\ & = 2^{24 - 4r}(x^{16 - 3r}) \end{align}

(i)(a)

\begin{align} 2^{x + 1} + y^2(y^{x - 2}) & = 11 \\ [a^{m + n} = (a^m)(a^n] \phantom{00000000} (2^x)(2^1) + + y^2(y^{x - 2}) & = 11 \\ \left[ a^{m - n} = {a^m \over a^n} \right] \phantom{00000000} (2^x)(2^1) + y^2\left( y^x \over y^2 \right) & = 11 \\ 2(2^x) + y^x & = 11 \\ 2(u) + (v) & = 11 \\ 2u + v & = 11 \\ \\ v & = 11 - 2u \phantom{0} \text{ --- (1)} \end{align}

(i)(b)

\begin{align} 4^{{1 \over 2}x + 1} + 5^{-1}y^x & = 4 \\ (2^2)^{{1 \over 2}x + 1} + 5^{-1}y^x & = 4 \\ [ (a^m)^n = a^{mn} ] \phantom{00000000} 2^{2({1 \over 2}x + 1)} + \left(1 \over 5 \right)y^x & = 4 \\ 2^{x + 2} + {y^x \over 5} & = 4 \\ [ a^{m + n} = (a^m)(a^n) ] \phantom{00000000} (2^x)(2^2) + {y^x \over 5} & = 4 \\ (2^x)(4) + {y^x \over 5} & = 4 \\ 4(2^x) + {y^x \over 5} & = 4 \\ 4(u) + {v \over 5} & = 4 \\ 4u + {1 \over 5}v & = 4 \\ \\ 20u + v & = 20 \phantom{0} \text{ --- (2)} \end{align}

(ii)

\begin{align} \text{Substitute } & \text{(1) into (2),} \\ 20u + (11 - 2u) & = 20 \\ 20u + 11 - 2u & = 20 \\ 18u & = 9 \\ u & = {9 \over 18} \\ & = {1 \over 2} \\ \\ \text{Substitute } & u = {1 \over 2} \text{ into (1),} \\ v & = 11 - 2\left(1 \over 2\right) \\ & = 10 \end{align}

(iii)

\begin{align} (4y)^x & = (4^x) (y^x) \\ & = (2^2)^x(y^x) \\ & = (2^{2x})(y^x) \phantom{00000000} [ (a^m)^n = a^{mn}] \\ & = (2^x)^2 (y^x) \\ & = (u)^2 (v) \\ & = \left(1 \over 2\right)^2 (10) \\ & = {5 \over 2} \end{align}

(i)

\begin{align} a^x & = b^y \\ a & = \sqrt[x]{b^y} \\ a & = b^{y \over x} \\ \\ \\ b^y & = (ab)^z \\ b^y & = a^z b^z \\ \\ \text{Since } & a = b^{y \over x}, \\ b^y & = \left(b^{y \over x}\right)^z b^z \\ b^y & = b^{zy \over x} b^z \\ b^y & = b^{ {zy \over x} + z} \\ \\ y & = {zy \over x} + z \\ x(y) & = x \left( {zy \over x} + z \right) \\ xy & = zy + zx \\ xy & = z(y + x) \\ {xy \over y + x} & = z \phantom{00} \text{ (Shown)} \end{align}

(ii)

\begin{align} a^x & = b^y \\ a & = \sqrt[x]{b^y} \\ a & = b^{y \over x} \\ \\ \\ b^y & = \left(a \over b\right)^w \\ b^y & = {a^w \over b^w} \\ b^w (b^y) & = a^w \\ b^{w + y} & = a^w \\ \ \text{Since } & a = b^{y \over x}, \\ b^{w + y} & = \left(b^{y \over x}\right)^w \\ b^{w + y} & = b^{wy \over x} \\ \\ w + y & = {wy \over x} \\ x(w + y) & = wy \\ wx + xy & = wy \\ xy & = wy - wx \\ xy & = w(y - x) \\ {xy \over y - x} & = w \phantom{00} \text{ (Shown)} \end{align}