Question 1: Assuming that $\displaystyle x$ is a positive real number and $\displaystyle a, b, c$ are rational numbers, show:

$\displaystyle \text{(i) } \bigg( \frac{x^b}{x^c} \bigg)^a . \bigg( \frac{x^c}{x^a} \bigg)^b . \bigg( \frac{x^a}{x^b} \bigg)^c = 1$

$\displaystyle \text{(ii) } \bigg( \frac{x^a}{x^b} \bigg)^{\frac{1}{ab}} . \bigg( \frac{x^b}{x^c} \bigg)^{\frac{1}{bc}} . \bigg( \frac{x^c}{x^a} \bigg)^{\frac{1}{ac}} = 1$

$\displaystyle \text{(iii) } \bigg( \frac{x^a}{x^b} \bigg)^{a^2+ab+b^2} . \bigg( \frac{x^b}{x^c} \bigg)^{b^2+bc+c^2} . \bigg( \frac{x^c}{x^a} \bigg)^{c^2+ca+a^2} = 1$

$\displaystyle \text{(iv) } \bigg( \frac{x^a}{x^b} \bigg)^{a+b} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a} = 1$

$\displaystyle \text{(v) } \bigg( \frac{x^a}{x^b} \bigg)^{a+b-c} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c-a} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a-b} = 1$

$\displaystyle \text{(vi) } \bigg( \frac{x^a}{x^{-b}} \bigg)^{a^2-ab+b^2} . \bigg( \frac{x^b}{x^{-c}} \bigg)^{b^2-bc+c^2} . \bigg( \frac{x^c}{x^{-a}} \bigg)^{c^2-ca+a^2} = x^{2(a^3 + b^3 + c^3)}$

$\displaystyle \text{(vii) } \bigg( \frac{x^{a(b-c)}}{x^{b(a-c)}} \bigg) \div \bigg( \frac{x^b}{x^a} \bigg)^c = 1$

$\displaystyle \text{(viii) } \frac{(x^{a+b})^2 . (x^{b+c})^2 . (x^{c+a})^2}{(x^a.x^b.x^c)^4} = 1$

$\displaystyle \text{(ix) } \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{a-b}+x^{c-b}} + \frac{1}{1+x^{b-c}+x^{a-c}} = 1$

$\displaystyle \text{(x) } \frac{a^{-1}}{a^{-1}+ b^{-1}} + \frac{a^{-1}}{a^{-1}- b^{-1}} = \frac{2b^2}{b^2-a^2}$

$\displaystyle \text{(xi) } abc=1 \text{ , show that } \frac{1}{1+a+b^{-1}}+ \frac{1}{1+b+c^{-1}} + \frac{1}{1+c+a^{-1}} = 1$

$\displaystyle \text{(xii) } \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1$

$\displaystyle \text{(xiii) } \frac{a+b+c}{a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$

$\displaystyle \text{(xiv) } (a^{-1}+b^{-1})^{-1} = \frac{ab}{a+b}$

$\displaystyle \text{(i) } \bigg( \frac{x^b}{x^c} \bigg)^a . \bigg( \frac{x^c}{x^a} \bigg)^b . \bigg( \frac{x^a}{x^b} \bigg)^c = 1$

$\displaystyle \bigg( \frac{x^b}{x^c} \bigg)^a . \bigg( \frac{x^c}{x^a} \bigg)^b . \bigg( \frac{x^a}{x^b} \bigg)^c$

$\displaystyle = (x^{b-c})^a . (x^{c-a})^b. (x^{a-b})^c$

$\displaystyle = x^{ab-ac}.x^{bc-ab}.x^{ac-bc}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(ii) } \bigg( \frac{x^a}{x^b} \bigg)^{\frac{1}{ab}} . \bigg( \frac{x^b}{x^c} \bigg)^{\frac{1}{bc}} . \bigg( \frac{x^c}{x^a} \bigg)^{\frac{1}{ac}} = 1$

$\displaystyle \bigg( \frac{x^a}{x^b} \bigg)^{\frac{1}{ab}} . \bigg( \frac{x^b}{x^c} \bigg)^{\frac{1}{bc}} . \bigg( \frac{x^c}{x^a} \bigg)^{\frac{1}{ac}}$

$\displaystyle = (x^{a-b})^{\frac{1}{ab}}.(x^{b-c})^{\frac{1}{bc}}.(x^{c-a})^{\frac{1}{ac}}$

$\displaystyle = x^{\frac{a-b}{ab}}. x^{\frac{b-c}{bc}}. x^{\frac{c-a}{ac}}$

$\displaystyle = x^{\frac{1}{b} -\frac{1}{a}}.x^{\frac{1}{c} -\frac{1}{b}}.x^{\frac{1}{a} -\frac{1}{c}}$

$\displaystyle = x^{\frac{1}{b} -\frac{1}{a}+\frac{1}{c} -\frac{1}{b}+\frac{1}{a} -\frac{1}{c}}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(iii) } \bigg( \frac{x^a}{x^b} \bigg)^{a^2+ab+b^2} . \bigg( \frac{x^b}{x^c} \bigg)^{b^2+bc+c^2} . \bigg( \frac{x^c}{x^a} \bigg)^{c^2+ca+a^2} = 1$

$\displaystyle \bigg( \frac{x^a}{x^b} \bigg)^{a^2+ab+b^2} . \bigg( \frac{x^b}{x^c} \bigg)^{b^2+bc+c^2} . \bigg( \frac{x^c}{x^a} \bigg)^{c^2+ca+a^2}$

$\displaystyle = (x^{a-b})^{a^2+ab+b^2} . (x^{b-c})^{b^2+bc+c^2}. (x^{c-a})^{c^2+ca+a^2}$

$\displaystyle = x^{(a-b)(a^2+ab+b^2)} . x^{(b-c)(b^2+bc+c^2)} . x^{(c-a)(c^2+ca+a^2)}$

$\displaystyle = x^{a^3-b^3} . x^{b^3-c^3} . x^{c^3-a^3}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(iv) } \bigg( \frac{x^a}{x^b} \bigg)^{a+b} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a} = 1$

$\displaystyle \bigg( \frac{x^a}{x^b} \bigg)^{a+b} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a}$

$\displaystyle = (x^{a-b})^{a+b} . (x^{b-c})^{b+c} . (x^{c-a})^{c+a}$

$\displaystyle = x^{(a-b)(a+b)} . x^{(b-c)(b+c)} . x^{(c-a)(c+a)}$

$\displaystyle = x^{a^2 - b^2} . x^{b^2 - c^2} . x^{c^2 - a^2}$

$\displaystyle = x^{a^2 - b^2 + b^2 - c^2 + c^2 - a^2}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(v) } \bigg( \frac{x^a}{x^b} \bigg)^{a+b-c} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c-a} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a-b} = 1$

$\displaystyle \bigg( \frac{x^a}{x^b} \bigg)^{a+b-c} . \bigg( \frac{x^b}{x^c} \bigg)^{b+c-a} . \bigg( \frac{x^c}{x^a} \bigg)^{c+a-b}$

$\displaystyle = (x^{a-b})^{a+b-c} . (x^{b-c})^{b+c-a} . (x^{c-a})^{c+a-b}$

$\displaystyle = x^{(a-b)(a+b-c)} . x^{(b-c)(b+c-a)} . x^{(c-a)(c+a-b)}$

$\displaystyle = x^{(a-b)(a+b)-(a-b)c)} . x^{(b-c)(b+c) - (b-c)a)} . x^{(c-a)(c+a) -(c-a)b)}$

$\displaystyle = x^{a^2-b^2-ac+bc} . x^{b^2-c^2-ba+ca} . x^{c^2-a^2-cb+ab}$

$\displaystyle = x^{a^2-b^2-ac+bc + b^2-c^2-ba+ca + c^2-a^2-cb+ab}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(vi) } \bigg( \frac{x^a}{x^{-b}} \bigg)^{a^2-ab+b^2} . \bigg( \frac{x^b}{x^{-c}} \bigg)^{b^2-bc+c^2} . \bigg( \frac{x^c}{x^{-a}} \bigg)^{c^2-ca+a^2} = x^{2(a^3 + b^3 + c^3)}$

$\displaystyle \bigg( \frac{x^a}{x^{-b}} \bigg)^{a^2-ab+b^2} . \bigg( \frac{x^b}{x^{-c}} \bigg)^{b^2-bc+c^2} . \bigg( \frac{x^c}{x^{-a}} \bigg)^{c^2-ca+a^2}$

$\displaystyle = (x^{a+b})^{a^2-ab+b^2} . (x^{b+c})^{b^2-bc+c^2} . (x^{c+a})^{c^2-ca+a^2}$

$\displaystyle = x^{(a+b)(a^2-ab+b^2)+ (b+c)(b^2-bc+c^2) + (c+a)(c^2-ca+a^2)}$

$\displaystyle = x^{a^3+b^3} . x^{b^3+c^3} . x^{c^3+a^3}$

$\displaystyle = x^{2(a^3 + b^3 + c^3)}$

$\displaystyle \\$

$\displaystyle \text{(vii) } \bigg( \frac{x^{a(b-c)}}{x^{b(a-c)}} \bigg) \div \bigg( \frac{x^b}{x^a} \bigg)^c = 1$

$\displaystyle \bigg( \frac{x^{a(b-c)}}{x^{b(a-c)}} \bigg) \div \bigg( \frac{x^b}{x^a} \bigg)^c$

$\displaystyle = \frac{ab-ac}{ba-bc} \div (\frac{x^b}{x^a})^c$

$\displaystyle = x^{ab-ac-ba+bc} . \frac{1}{x^{(b-a)c}}$

$\displaystyle = x^{ab-ac-ba+bc} . \frac{1}{x^{bc-ac}}$

$\displaystyle = x^{-ac+bc} . x^{ac-bc}$

$\displaystyle = x^{-ac+bc+ac-bc}$

$\displaystyle = x^0 = 1$

$\displaystyle \\$

$\displaystyle \text{(viii) } \frac{(x^{a+b})^2 . (x^{b+c})^2 . (x^{c+a})^2}{(x^a.x^b.x^c)^4} = 1$

$\displaystyle \frac{(x^{a+b})^2 . (x^{b+c})^2 . (x^{c+a})^2}{(x^a.x^b.x^c)^4}$

$\displaystyle = \frac{x^{2(a+b)} . x^{2(b+c)} . (x^{2(c+a)}}{(x^a.x^b.x^c)^4}$

$\displaystyle = \frac{x^{2a+2b} . x^{2b+2c} . (x^{2c+2a}}{x^{4a}.x^{4b}.x^{4c}}$

$\displaystyle = \frac{x^{2a+2b+2b+2c+2c+2a}}{x^{4a+4b+4c}}$

$\displaystyle = \frac{x^{4a+4b+4c}}{x^{4a+4b+4c}}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(ix) } \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{a-b}+x^{c-b}} + \frac{1}{1+x^{b-c}+x^{a-c}} = 1$

$\displaystyle \frac{1}{1+x^{b-a}+x^{c-a}} + \frac{1}{1+x^{a-b}+x^{c-b}} + \frac{1}{1+x^{b-c}+x^{a-c}}$

$\displaystyle = \frac{x^a}{x^a+x^{b-a+a}+x^{c-a+a}} + \frac{x^b}{x^b+x^{a-b+b}+x^{c-b+b}} + \frac{x^c}{x^c+x^{b-c+c}+x^{a-c+c}}$

$\displaystyle = \frac{x^a}{x^a+x^{b}+x^{c}} + \frac{x^b}{x^b+x^{a}+x^{c}} + \frac{x^c}{x^c+x^{b}+x^{a}}$

$\displaystyle = \frac{x^a+x^b+x^c}{x^a+x^b+x^c}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(x) } \frac{a^{-1}}{a^{-1}+ b^{-1}} + \frac{a^{-1}}{a^{-1}- b^{-1}} = \frac{2b^2}{b^2-a^2}$

$\displaystyle \frac{a^{-1}}{a^{-1}+ b^{-1}} + \frac{a^{-1}}{a^{-1}- b^{-1}}$

$\displaystyle = \frac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}} + \frac{\frac{1}{a}}{\frac{1}{a}-\frac{1}{b}}$

$\displaystyle = \frac{\frac{1}{a}}{\frac{a+b}{ab}} + \frac{\frac{1}{a}}{\frac{b-a}{ab}}$

$\displaystyle = \frac{1}{a} . \frac{ab}{a+b} + \frac{1}{a} . \frac{ab}{b-a}$

$\displaystyle = \frac{b}{a+b}+ \frac{b}{b-a}$

$\displaystyle = \frac{b(b-a)+b(a+b)}{(a+b)(b-a)}$

$\displaystyle = \frac{b^2 - ba + ba + b^2}{b^2-a^2}$

$\displaystyle = \frac{2b^2}{b^2-a^2}$

$\displaystyle \\$

(xi) If $\displaystyle abc=1$, show that $\displaystyle \frac{1}{1+a+b^{-1}}+ \frac{1}{1+b+c^{-1}} + \frac{1}{1+c+a^{-1}} = 1$

$\displaystyle \frac{1}{1+a+b^{-1}}+ \frac{1}{1+b+c^{-1}} + \frac{1}{1+c+a^{-1}}$

$\displaystyle = \frac{1}{1+a+\frac{1}{b}}+ \frac{1}{1+b+\frac{1}{c}} + \frac{1}{1+c+\frac{1}{a}}$

Given $\displaystyle abc=1 \Rightarrow c = \frac{1}{ab}$

$\displaystyle = \frac{1}{1+a+\frac{1}{b}}+ \frac{1}{1+b+ab} + \frac{1}{1+\frac{1}{ab}+\frac{1}{a}}$

$\displaystyle = \frac{b}{b+ab+1} + \frac{1}{b+ab+1} + \frac{ab}{b+ab+1}$

$\displaystyle = \frac{b+1+ab}{b+1+ab}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(xii) } \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1$

$\displaystyle \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}}$

$\displaystyle = \frac{1}{1+\frac{x^a}{x^b}} + \frac{1}{1+\frac{x^b}{x^a}}$

$\displaystyle = \frac{x^b}{x^a+x^b} + \frac{x^a}{x^a+x^b}$

$\displaystyle = \frac{x^a + x^b}{x^a+x^b}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(xiii) } \frac{a+b+c}{a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}} = abc$

$\displaystyle \frac{a+b+c}{a^{-1}b^{-1} + b^{-1}c^{-1} + c^{-1}a^{-1}}$

$\displaystyle = \frac{a+b+c}{\frac{1}{ab} + \frac{1}{bc} + \frac{1}{ac}}$

$\displaystyle = \frac{abc(a+b+c)}{(a+b+c)}$

$\displaystyle = abc$

$\displaystyle \\$

$\displaystyle \text{(xiv) } (a^{-1}+b^{-1})^{-1} = \frac{ab}{a+b}$

$\displaystyle (a^{-1}+b^{-1})^{-1}$

$\displaystyle = \frac{1}{\frac{1}{a}+\frac{1}{b}}$

$\displaystyle = \frac{ab}{a+b}$

$\displaystyle \\$

Question 2: Assuming $\displaystyle x, y, z$ are positive real numbers, simplify the following:

$\displaystyle \text{(i) } \sqrt{x^{-2}y^3}$          $\displaystyle \text{(ii) } (x^{\frac{-2}{3}})^2 (y^{\frac{-1}{2}})^2$           $\displaystyle \text{(iii) } (\sqrt{x^{-3}})^5$

$\displaystyle \text{(iv) } (\sqrt{x})^{-2/3} \sqrt{y^4} \div \sqrt{xy^{-1/2}}$          $\displaystyle \text{(v) } \sqrt[3]{xy^2} \div x^2y$

$\displaystyle \text{(vi) } \sqrt[4]{\sqrt[3]{x^2}}$           $\displaystyle \text{(vii) } \bigg( \frac{x^{a+b}}{x^c} \bigg)^{a-b} . \bigg( \frac{x^{b+c}}{x^a} \bigg)^{b-c} . \bigg( \frac{x^{c+a}}{x^b} \bigg)^{c-a}$

$\displaystyle \text{(viii) } \sqrt[lm]{\frac{x^l}{x^m}} . \sqrt[mn]{\frac{x^m}{x^n}} . \sqrt[nl]{\frac{x^n}{x^l}}$

$\displaystyle \text{(i) } \sqrt{x^{-2}y^3}$

$\displaystyle \sqrt{x^{-2}y^3} = \sqrt{\frac{y^3}{x^2}} = (\frac{y^3}{x^2})^{\frac{1}{2}} = \frac{y^{\frac{3}{2}}}{x^{\frac{2}{2}}} = \frac{y^{\frac{3}{2}}}{x}$

$\displaystyle \\$

$\displaystyle \text{(ii) } (x^{\frac{-2}{3}})^2 (y^{\frac{-1}{2}})^2$

$\displaystyle (x^{\frac{-2}{3}})^2 (y^{\frac{-1}{2}})^2 = x^{\frac{-4}{3}} y^{-1} = \frac{1}{x^{ \frac{4}{3}} y}$

$\displaystyle \\$

$\displaystyle \text{(iii) } (\sqrt{x^{-3}})^5$

$\displaystyle (\sqrt{x^{-3}})^5 = (x^{\frac{-3}{2}})^5 = x^{\frac{-15}{2}} = \frac{1}{x^{\frac{15}{2}}}$

$\displaystyle \\$

$\displaystyle \text{(iv) } (\sqrt{x})^{-2/3} \sqrt{y^4} \div \sqrt{xy^{-1/2}}$

$\displaystyle (\sqrt{x})^{-2/3} \sqrt{y^4} \div \sqrt{xy^{-1/2}} = \frac{ (x^{\frac{1}{2}})^{\frac{-2}{3}} (y^4)^{\frac{1}{2}} }{(xy^{\frac{-1}{2}})^{\frac{1}{2}}} = \frac{x^{\frac{-1}{3}} y^2}{x^{\frac{1}{2}} y^{\frac{-1}{4}}} = \frac{y^{\frac{9}{8}}}{x^{\frac{5}{6}}}$

$\displaystyle \\$

$\displaystyle \text{(v) } \sqrt[3]{xy^2} \div x^2y$

$\displaystyle \sqrt[3]{xy^2} \div x^2y = \frac{ (xy^2)^{\frac{1}{3}} } {x^2 y} = \frac{ x^{\frac{1}{3}} y^{\frac{2}{3}} }{ x^2y} = x^{\frac{-5}{3}} y^{\frac{-1}{3}} = \frac{1}{ x^{\frac{5}{3}} y^{\frac{1}{3}}}$

$\displaystyle \\$

$\displaystyle \text{(vi) } \sqrt[4]{\sqrt[3]{x^2}}$

$\displaystyle \sqrt[4]{\sqrt[3]{x^2}} = \{ (x^2)^{\frac{1}{3}} \}^{\frac{1}{4}} = (x^{\frac{2}{3}})^{\frac{1}{4}} = x^{\frac{1}{6}}$

$\displaystyle \\$

$\displaystyle \text{(vii) } \bigg( \frac{x^{a+b}}{x^c} \bigg)^{a-b} . \bigg( \frac{x^{b+c}}{x^a} \bigg)^{b-c} . \bigg( \frac{x^{c+a}}{x^b} \bigg)^{c-a}$

$\displaystyle \bigg( \frac{x^{a+b}}{x^c} \bigg)^{a-b} . \bigg( \frac{x^{b+c}}{x^a} \bigg)^{b-c} . \bigg( \frac{x^{c+a}}{x^b} \bigg)^{c-a}$

$\displaystyle = \bigg( \frac{x^{a^2-b^2}}{x^{ca - cb}} \bigg) . \bigg( \frac{x^{b^2-c^2}}{x^{ab - ac}} \bigg) . \bigg( \frac{x^{c^2-a^2}}{x^{bc - ba}} \bigg)$

$\displaystyle = \frac{x^{a^2-b^2+b^2-c^2+c^2-a^2}}{x^{ca - cb+ab - ac+bc - ba}} = \frac{x^0}{x^0} = 1$

$\displaystyle \\$

$\displaystyle \text{(viii) } \sqrt[lm]{\frac{x^l}{x^m}} . \sqrt[mn]{\frac{x^m}{x^n}} . \sqrt[nl]{\frac{x^n}{x^l}}$

$\displaystyle \sqrt[lm]{\frac{x^l}{x^m}} . \sqrt[mn]{\frac{x^m}{x^n}} . \sqrt[nl]{\frac{x^n}{x^l}}$

$\displaystyle = (\frac{x^l}{x^m})^{\frac{1}{lm}} . (\frac{x^m}{x^n})^{\frac{1}{mn}} . (\frac{x^n}{x^l})^{\frac{1}{nl}}$

$\displaystyle = \frac{ x^{ \frac{1}{m}+ \frac{1}{n} + \frac{1}{l} }} { x^{ \frac{1}{l} + \frac{1}{m} + \frac{1}{ln} } } = 1$

$\displaystyle \\$

Question 3: Assuming $\displaystyle x, y, z$ are positive real numbers, show that:

$\displaystyle \text{(i) } \sqrt{x^{-1}y} \times \sqrt{y^{-1}z} \times \sqrt{z^{-1}x} = 1$

$\displaystyle \text{(ii) } \bigg( \frac{x^{-1}y^2}{x^3y^{-2}} \bigg)^{1/3} \div \bigg( \frac{x^6y^{-3}}{x^{-2}y^3} \bigg)^{1/2} = x^ay^b \text{ , prove that } a+b=-1 \text{ , where } x \text{ and } y \text{ are different. }$

$\displaystyle \text{(iii) } \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1$

$\displaystyle \text{(iv) } \bigg\{ \bigg( \frac{x^{a(a-b)}}{x^{a(a+b)}} \bigg) \div \bigg( \frac{x^{b(b-a)}}{x^{b(b+a)}} \bigg) \bigg\}^{(a+b)} = 1$

$\displaystyle \text{(v) } \bigg( x^{\frac{1}{a-b}} \bigg)^{\frac{1}{a-c}} . \bigg( x^{\frac{1}{b-c}} \bigg)^{\frac{1}{b-a}} . \bigg( x^{\frac{1}{c-a}} \bigg)^{\frac{1}{c-b}} = 1$

$\displaystyle \text{(vi) } \bigg( \frac{x^{a^2+b^2}}{x^{ab}} \bigg)^{a+b} . \bigg( \frac{x^{b^2+c^2}}{x^{bc}} \bigg)^{b+c} . \bigg( \frac{x^{c^2+a^2}}{x^{ac}} \bigg)^{a+c} = x^{2(a^3+b^3+c^3)}$

$\displaystyle \text{(vii) } (x^{a-b})^{a+b} . (x^{b-c})^{b+c} . (x^{c-a})^{c+a} = 1$

$\displaystyle \text{(viii) } \big\{ \big( x^{a- a^{-1}} \big)^{\frac{1}{a-1}} \big\}^{\frac{a}{a+1}} = x$

$\displaystyle \text{(ix) } \bigg( \frac{a^{x+1}}{a^{y+1}} \bigg)^{x+y} . \bigg( \frac{a^{y+2}}{a^{z+2}} \bigg)^{y+z} . \bigg( \frac{a^{z+3}}{a^{x+3}} \bigg)^{z+x} = 1$

$\displaystyle \text{(x) } \bigg( \frac{3^a}{3^b} \bigg)^{a+b} . \bigg( \frac{3^b}{3^c} \bigg)^{b+c} . \bigg( \frac{3^c}{3^a} \bigg)^{c+a} = 1$

$\displaystyle \text{(xi) } \frac{\big( a+ \frac{1}{b} \big)^m \times \big( a - \frac{1}{b} \big)^n}{\big( b+ \frac{1}{a} \big)^m \times \big( b - \frac{1}{a} \big)^n}= \big( \frac{a}{b} \big)^{m+n}$

$\displaystyle \text{(i) } \sqrt{x^{-1}y} \times \sqrt{y^{-1}z} \times \sqrt{z^{-1}x} = 1$

$\displaystyle \sqrt{x^{-1}y} \times \sqrt{y^{-1}z} \times \sqrt{z^{-1}x}$

$\displaystyle = (\frac{y}{x})^{\frac{1}{2}} \times (\frac{z}{y})^{\frac{1}{2}} \times (\frac{x}{z})^{\frac{1}{2}}$

$\displaystyle = 1$

$\displaystyle \\$

(ii) If $\displaystyle \bigg( \frac{x^{-1}y^2}{x^3y^{-2}} \bigg)^{1/3} \div \bigg( \frac{x^6y^{-3}}{x^{-2}y^3} \bigg)^{1/2} = x^ay^b$, prove that $\displaystyle a+b=-1$, where $\displaystyle x and y$ are different.

$\displaystyle \bigg( \frac{x^{-1}y^2}{x^3y^{-2}} \bigg)^{1/3} \div \bigg( \frac{x^6y^{-3}}{x^{-2}y^3} \bigg)^{1/2} = x^ay^b$

$\displaystyle \Rightarrow (x^{-4} y^4)^{\frac{1}{3}} \div (x^8y^{-6})^{\frac{1}{2}} = x^ay^b$

$\displaystyle \Rightarrow \frac{ x^{-4}{3} y^{\frac{4}{3}} } { x^4y^{-3}} = x^ay^b$

$\displaystyle \Rightarrow x^{\frac{-4}{3}-4}y^{\frac{4}{3}+3} = x^ay^b$

$\displaystyle \Rightarrow x^{\frac{-16}{3}}y^{\frac{13}{3} } = x^ay^b$

$\displaystyle \Rightarrow a= \frac{-16}{3} and b = \frac{13}{3}$

$\displaystyle \therefore a+b = \frac{-16}{3} + b= \frac{13}{3} = \frac{-3}{3} = -1$

$\displaystyle \\$

$\displaystyle \text{(iii) } \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}} = 1$

$\displaystyle \frac{1}{1+x^{a-b}} + \frac{1}{1+x^{b-a}}$

$\displaystyle = \frac{1}{1+\frac{x^a}{x^b}} + \frac{1}{1+\frac{x^b}{x^a}}$

$\displaystyle = \frac{x^b}{x^a+x^b} + \frac{x^a}{x^b+x^a}$

$\displaystyle = \frac{x^a+x^b}{x^a+x^b}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(iv) } \bigg\{ \bigg( \frac{x^{a(a-b)}}{x^{a(a+b)}} \bigg) \div \bigg( \frac{x^{b(b-a)}}{x^{b(b+a)}} \bigg) \bigg\}^{(a+b)} = 1$

$\displaystyle \bigg\{ \bigg( \frac{x^{a(a-b)}}{x^{a(a+b)}} \bigg) \div \bigg( \frac{x^{b(b-a)}}{x^{b(b+a)}} \bigg) \bigg\}^{(a+b)}$

$\displaystyle = \frac{x^{a(a^2-b^2)}}{x^{a(a+b)^2}} \times \frac{x^{b(a+b)^2}}{x^{b(b^2-a^2)}}$

$\displaystyle = \frac{x^{a^3-ab^2+ba^2+b^3+2ab^2}}{x^{a^3+ab^2+2a^b+b^3-ba^2}}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(v) } \bigg( x^{\frac{1}{a-b}} \bigg)^{\frac{1}{a-c}} . \bigg( x^{\frac{1}{b-c}} \bigg)^{\frac{1}{b-a}} . \bigg( x^{\frac{1}{c-a}} \bigg)^{\frac{1}{c-b}} = 1$

$\displaystyle \bigg( x^{\frac{1}{a-b}} \bigg)^{\frac{1}{a-c}} . \bigg( x^{\frac{1}{b-c}} \bigg)^{\frac{1}{b-a}} . \bigg( x^{\frac{1}{c-a}} \bigg)^{\frac{1}{c-b}}$

$\displaystyle = x^{\frac{1}{(a-b)(a-c)}} \times x^{\frac{1}{(b-c)(b-a)}} \times x^{\frac{1}{(c-a)(c-b)}}$

$\displaystyle = x^{\frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)}} \times x^{\frac{1}{(c-a)(c-b)}}$

$\displaystyle = x^{\frac{b^2-bc-ab+ac+a^2-ab-ac+bc}{(a-b)(a-c)(b-c)(b-a)}} \times x^{\frac{1}{(c-a)(c-b)}}$

$\displaystyle = x^{\frac{-1}{(a-c)(b-c)}} \times x^{\frac{1}{(c-a)(c-b)}}$

$\displaystyle = x^{\frac{-c^2+ac+bc-ab+ab-bc-ac+c^2}{(a-c)(b-c)(c-a)(c-b)}}$

$\displaystyle = x^0$

$\displaystyle \\$

$\displaystyle \text{(vi) } \bigg( \frac{x^{a^2+b^2}}{x^{ab}} \bigg)^{a+b} . \bigg( \frac{x^{b^2+c^2}}{x^{bc}} \bigg)^{b+c} . \bigg( \frac{x^{c^2+a^2}}{x^{ac}} \bigg)^{a+c} = x^{2(a^3+b^3+c^3)}$

$\displaystyle \bigg( \frac{x^{a^2+b^2}}{x^{ab}} \bigg)^{a+b} . \bigg( \frac{x^{b^2+c^2}}{x^{bc}} \bigg)^{b+c} . \bigg( \frac{x^{c^2+a^2}}{x^{ac}} \bigg)^{a+c}$

$\displaystyle = x^{(a^2+b^2-ab)(a+b)} . x^{(b^2+c^2-bc)(b+c)} . x^{(c^2+a^2-ac)(a+c)}$

$\displaystyle = x^{a^3+b^3}. x^{b^3+c^3}. x^{c^3+a^3}$

$\displaystyle = x^{2(a^3+b^3+c^3)}$

$\displaystyle \\$

$\displaystyle \text{(vii) } (x^{a-b})^{a+b} . (x^{b-c})^{b+c} . (x^{c-a})^{c+a} = 1$

$\displaystyle (x^{a-b})^{a+b} . (x^{b-c})^{b+c} . (x^{c-a})^{c+a}$

$\displaystyle = x^{a^2-b^2}.x^{b^2-c^2}.x^{c^2-a^2}$

$\displaystyle = x^0 =1$

$\displaystyle \\$

$\displaystyle \text{(viii) } \big\{ \big( x^{a- a^{-1}} \big)^{\frac{1}{a-1}} \big\}^{\frac{a}{a+1}} = x$

$\displaystyle \big\{ \big( x^{a- a^{-1}} \big)^{\frac{1}{a-1}} \big\}^{\frac{a}{a+1}}$

$\displaystyle = \big\{ \big( x^{\frac{a^2-1}{a}} \big)^{\frac{1}{a-1}} \big\}^{\frac{a}{a+1}}$

$\displaystyle = \big\{ \big( x^{\frac{(a+1)(a-1)}{a}} \big)^{\frac{1}{a-1}} \big\}^{\frac{a}{a+1}}$

$\displaystyle = x^{\frac{(a+1)(a-1)}{a} \times \frac{1}{a-1} \times \frac{a}{a+1} }$

$\displaystyle = x^1 = x$

$\displaystyle \\$

$\displaystyle \text{(ix) } \bigg( \frac{a^{x+1}}{a^{y+1}} \bigg)^{x+y} . \bigg( \frac{a^{y+2}}{a^{z+2}} \bigg)^{y+z} . \bigg( \frac{a^{z+3}}{a^{x+3}} \bigg)^{z+x} = 1$

$\displaystyle \bigg( \frac{a^{x+1}}{a^{y+1}} \bigg)^{x+y} . \bigg( \frac{a^{y+2}}{a^{z+2}} \bigg)^{y+z} . \bigg( \frac{a^{z+3}}{a^{x+3}} \bigg)^{z+x}$

$\displaystyle = \frac{a^{(x+1)(x+y)} . a^{(y+2)(y+z)} . a^{(z+3)(z+x)}}{a^{(y+1)(x+y)}. a^{(z+2)(y+z)}. a^{(x+3)(z+x)}}$

$\displaystyle = \frac{a^{x^2+x+xy+y+y^2+2y+yz+2z+z^2+3z+zx+3x}}{a^{y^2+xy+x+y+zy+2y+z^2+2z+xz+3z+x^2+3x}}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(x) } \bigg( \frac{3^a}{3^b} \bigg)^{a+b} . \bigg( \frac{3^b}{3^c} \bigg)^{b+c} . \bigg( \frac{3^c}{3^a} \bigg)^{c+a} = 1$

$\displaystyle \bigg( \frac{3^a}{3^b} \bigg)^{a+b} . \bigg( \frac{3^b}{3^c} \bigg)^{b+c} . \bigg( \frac{3^c}{3^a} \bigg)^{c+a}$

$\displaystyle = \frac{3^{a(a+b)+b(b+c)+c(c+a)}}{3^{b(a+b)+c(b+c)+a(c+a)}}$

$\displaystyle = \frac{3^{a^2+ab+b^2+bc+c^2+ac}}{3^{b^2+ab+c^2+bc+a^2+ac}}$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(xi) } \frac{\big( a+ \frac{1}{b} \big)^m \times \big( a - \frac{1}{b} \big)^n}{\big( b+ \frac{1}{a} \big)^m \times \big( b - \frac{1}{a} \big)^n}= \big( \frac{a}{b} \big)^{m+n}$

$\displaystyle \frac{\big( a+ \frac{1}{b} \big)^m \times \big( a - \frac{1}{b} \big)^n}{\big( b+ \frac{1}{a} \big)^m \times \big( b - \frac{1}{a} \big)^n}$

$\displaystyle = \frac{(ab+1)^m a^m}{b^m (ab+1)^m} . \frac{(ab-1)^n a^n}{b^n (ab-1)^n}$

$\displaystyle = \frac{a^m}{b^m}. \frac{a^n}{b^n}$

$\displaystyle = (\frac{a}{b})^{m+n}$

$\displaystyle \\$

Question 4:

$\displaystyle \text{(i) } \text{If } a = x^{m+n}y^l, b= x^{n+l}y^m \text{ and } c = x^{l+m}y^n \text{, prove that } a^{m-n}b^{n-l}c^{l-m} = 1$

$\displaystyle \text{(ii) } \text{If } x = a^{m+n}, y = a^{n+l} \text{ and } z = a^{l+m} \text{, prove that } x^m y^n z^l = x^n y^l z^m$

$\displaystyle \text{(iii) } \text{If } a=xy^{p-1}, b= xy^{q-1} \text{ and } c= xy^{r-1} \text{, prove that } a^{q-r}b^{r-p}c^{p-q}=1$

$\displaystyle \text{(i) } \text{If } a = x^{m+n}y^l, b= x^{n+l}y^m \text{ and } c = x^{l+m}y^n \text{, prove that } a^{m-n}b^{n-l}c^{l-m} = 1$

$\displaystyle a^{m-n}n^{n-l}c^{l-m}$

$\displaystyle = {(x^{m+n}y^l)}^{m-n} .{(x^{n+l}y^m)}^{n-l} . {(x^{l+m}y^n)}^{l-m}$

$\displaystyle = x^{(m+n)(m-n)+(n+l)(n-l)+(l+m)(l-m)}. y^{l(m-n)+m(n-l)+n(l-m)}$

$\displaystyle = x^{m^2-n^2+n^2-l^2+l^2-m^2}.y^{lm-ln+mn-ml+nl-nm}$

$\displaystyle = x^0.y^0$

$\displaystyle = 1$

$\displaystyle \\$

$\displaystyle \text{(ii) } \text{If } x = a^{m+n}, y = a^{n+l} \text{ and } z = a^{l+m} \text{, prove that } x^m y^n z^l = x^n y^l z^m$

RHS $\displaystyle = x^m y^n z^l= {(a^{m+n})}^m {(a^{n+l})}^n {(a^{l+m})}^l$

$\displaystyle = a^{m^2+mn+n^2+nl+l^2+lm}$

$\displaystyle = a^{m^2+n^2+l^2+mn+nl+lm}$

LHS $\displaystyle = x^n y^l z^m = {(a^{m+n})}^n {(a^{n+l})}^l {(a^{l+m})}^m$

$\displaystyle = a^{mn+n^2+nl+l^2+lm+m^2}$

$\displaystyle = a^{m^2+n^2+l^2+mn+nl+lm}$

Therefore RHS = LHS. Hence proved.

$\displaystyle \\$

$\displaystyle \text{(iii) } \text{If } a=xy^{p-1}, b= xy^{q-1} \text{ and } c= xy^{r-1} \text{, prove that } a^{q-r}b^{r-p}c^{p-q}=1$

$\displaystyle (xy^{p-1})^{(q-r)}(xy^{q-1})^{(r-p)}(xy^{r-1})^{(p-q)}$

$\displaystyle = x^{q-r+r-p+p-q}. y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}$

$\displaystyle = x^0. y^{pq-q-rp+r+rq-r-pq+p+rp-p-rq+q}$

$\displaystyle = x^0.y^0$

$\displaystyle = 1$

$\displaystyle \\$

Question 5:

(i) If $\displaystyle a and b$ are distinct positive primes such that $\displaystyle \sqrt[3]{a^6b^{-4}}=a^xb^{2y}$, find $\displaystyle x and y$

(ii) If $\displaystyle a and b$ are distinct positive primes such that $\displaystyle (a+b)^{-1}(a^{-1}+b^{-1})=a^xb^y$ , find $\displaystyle x+y+2$

(i) If $\displaystyle a and b$ are distinct positive primes such that $\displaystyle \sqrt[3]{a^6b^{-4}}=a^xb^{2y}$, find $\displaystyle x and y$

$\displaystyle \sqrt[3]{a^6b^{-4}}=a^xb^{2y}$

$\displaystyle \Rightarrow (a^6.b^{-4})^{\frac{1}{3}}= a^xb^{2y}$

$\displaystyle \Rightarrow a^2b^{\frac{-4}{3}} = a^xb^{2y}$

$\displaystyle \Rightarrow x = 2 and y = \frac{-2}{3}$

$\displaystyle \\$

(ii) If $\displaystyle a and b$ are distinct positive primes such that $\displaystyle (a+b)^{-1}(a^{-1}+b^{-1})=a^xb^y$ , find $\displaystyle x+y+2$

$\displaystyle (a+b)^{-1}(a^{-1}+b^{-1})=a^xb^y$

$\displaystyle \Rightarrow \frac{1}{a+b} . (\frac{a+b}{ab}) = a^xb^y$

$\displaystyle \Rightarrow a^{-1}b^{-1} = a^xb^y$

Therefore $\displaystyle x = -1, y = -1$

Hence $\displaystyle x+y+2 = 0$

$\displaystyle \\$

Question 6: If $\displaystyle 2^x \times 3^y \times 5^z = 2160$, find $\displaystyle x, y and z$. Then compute the value of $\displaystyle 3^x \times 2^{-y} \times 5^{-z}$

$\displaystyle 2160 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 = 2^4 \times 3^3 \times 5^1$

$\displaystyle 2^x \times 3^y \times 5^z = 2^4 \times 3^3 \times 5^1$

Therefore $\displaystyle x = 4, y = 3 and z = 1$

Hence $\displaystyle 3^x \times 2^{-y} \times 5^{-z}$ = $\displaystyle 3^4 \times 2^{-3} \times 5^{-1} = \frac{81}{40}$

$\displaystyle \\$

Question 7:

(i) If $\displaystyle x= 2^{1/3} + 2^{2/3}$, show that $\displaystyle x^3-6x=6$

(ii) Determine $\displaystyle (8x)^x$, if $\displaystyle 9^{x+2}= 240+9^x$

(iii) If $\displaystyle 3^{4x}= (81)^{-1}$ and $\displaystyle 10^{1/y}= 0.0001$, find the value of $\displaystyle 2^{-x+4y}$

(i) If $\displaystyle x= 2^{1/3} + 2^{2/3}$, show that $\displaystyle x^3-6x=6$

We know: $\displaystyle (a+b)^3 = a^3 + b^3 + 3ab(a+b)$

Therefore $\displaystyle x^3 = (2^{\frac{1}{3}}+2^{\frac{2}{3}})^3$

$\displaystyle = 2+4+6(2^{\frac{1}{3}}+2^{\frac{2}{3}})$

$\displaystyle = 6+ 6(2^{\frac{1}{3}}+2^{\frac{2}{3}})$

Therefore $\displaystyle x^3 - 6x = 6+ 6(2^{\frac{1}{3}}+2^{\frac{2}{3}}) - 6(2^{\frac{1}{3}}+2^{\frac{2}{3}}) = 6$

Hence Proved.

$\displaystyle \\$

(ii) Determine $\displaystyle (8x)^x$, if $\displaystyle 9^{x+2}= 240+9^x$

$\displaystyle 9^{x+2}= 240+9^x$

$\displaystyle \Rightarrow 9^{x+2} - 9^x = 240$

$\displaystyle \Rightarrow 9^x(81-1) = 240$

$\displaystyle \Rightarrow 9^x = 3$

$\displaystyle \Rightarrow x = \frac{1}{2}$

Therefore $\displaystyle (8x)^x = (8 \times \frac{1}{2})^{\frac{1}{2}} = 4^{\frac{1}{2}} = 2$

$\displaystyle \\$

(iii) If $\displaystyle 3^{4x}= (81)^{-1}$ and $\displaystyle 10^{1/y}= 0.0001$, find the value of $\displaystyle 2^{-x+4y}$

$\displaystyle 3^{4x}= (81)^{-1}$

$\displaystyle \Rightarrow 3^{4x}= (3^4)^{-1}$

$\displaystyle \Rightarrow 4x = -4 or x = -1$

Similarly, $\displaystyle 10^{1/y}= 0.0001$

$\displaystyle \Rightarrow 10^{1/y}= (10)^{-4}$

$\displaystyle \Rightarrow y = \frac{-1}{4}$

Hence $\displaystyle 2^{-x+4y} \Rightarrow x^{1-1}=x^0 = 1$

$\displaystyle \\$

Question 8:

(i) If $\displaystyle a^x=b^y=c^z$ and $\displaystyle b^2 = ac$, then show that $\displaystyle y = \frac{2zx}{z+x}$

(ii) If $\displaystyle 2^x=3^y=6^{-z}$, show that $\displaystyle \frac{1}{x} + \frac{1}{y}+ \frac{1}{z} =0$

(iii) If $\displaystyle 2^x=3^y=12^z$, show that $\displaystyle \frac{1}{z} = \frac{1}{y}+ \frac{2}{x}$

(i) If $\displaystyle a^x=b^y=c^z$ and $\displaystyle b^2 = ac$, then show that $\displaystyle y = \frac{2zx}{z+x}$

Let $\displaystyle a^x=b^y=c^z = k$

Therefore $\displaystyle a = (k)^{\frac{1}{x}}$ , $\displaystyle b = (k)^{\frac{1}{y}}$ , $\displaystyle c = (k)^{\frac{1}{z}}$

Now, $\displaystyle b^2 = ac$

$\displaystyle \Rightarrow k^{\frac{2}{y}} = k^{\frac{1}{x}} . k^{\frac{1}{z}}$

$\displaystyle \Rightarrow \frac{2}{y} = \frac{1}{x}+\frac{1}{z}$

$\displaystyle \Rightarrow y = \frac{2xz}{x+z}$

$\displaystyle \\$

(ii) If $\displaystyle 2^x=3^y=6^{-z}$, show that $\displaystyle \frac{1}{x} + \frac{1}{y}+ \frac{1}{z} =0$

Let $\displaystyle 2^x=3^y=6^{-z}= k$

Therefore $\displaystyle 2 = (k)^{\frac{1}{x}}$ , $\displaystyle 3 = (k)^{\frac{1}{y}}$ , $\displaystyle 6 = (k)^{\frac{-1}{z}}$

$\displaystyle 6 = (k)^{\frac{-1}{z}}$

$\displaystyle \Rightarrow 2 \times 3 = (k)^{\frac{-1}{z}}$

$\displaystyle \Rightarrow (k)^{\frac{1}{x}} \times (k)^{\frac{1}{y}} = (k)^{\frac{-1}{z}}$

$\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y} = \frac{-1}{z}$

$\displaystyle \Rightarrow \frac{1}{x} + \frac{1}{y}+ \frac{1}{z} =0$

$\displaystyle \\$

(iii) If $\displaystyle 2^x=3^y=12^z$, show that $\displaystyle \frac{1}{z} = \frac{1}{y}+ \frac{2}{x}$

Let $\displaystyle 2^x=3^y=12^z = k$

Therefore $\displaystyle 2 = (k)^{\frac{1}{x}}$ , $\displaystyle 3 = (k)^{\frac{1}{y}}$ , $\displaystyle 12 = (k)^{\frac{1}{z}}$

$\displaystyle 12 = (k)^{\frac{1}{z}}$

$\displaystyle \Rightarrow 2^2 \times 3 = (k)^{\frac{1}{z}}$

$\displaystyle \Rightarrow ({(k)^{\frac{1}{x}}})^2 \times (k)^{\frac{1}{y}} = (k)^{\frac{1}{z}}$

$\displaystyle \Rightarrow \frac{2}{x}+\frac{1}{y} = \frac{1}{z}$

$\displaystyle \\$

Question 9: Solve the following equations:

$\displaystyle \text{(i) } 2^{x-5} = 256$

$\displaystyle \text{(ii) } 2^{x+3} = 4^{x-1}$

$\displaystyle \text{(iii) } 2^{2x+1} = 17. 2^x - 2^3$

$\displaystyle \text{(iv) } 5^{2x+1} = 6.5^x - 1$

$\displaystyle \text{(v) } 2^{2x} - 2^{x+3}+2^4 = 0$

$\displaystyle \text{(vi) } 3^{2x+4} + 1 = 2. 3^{x+2}$

$\displaystyle \text{(i) } 2^{x-5} = 256$

$\displaystyle 2^{x-5} = 256$

$\displaystyle 2^{x-5} = 2^8$

$\displaystyle \Rightarrow x-5 = 8$

$\displaystyle \Rightarrow x = 13$

$\displaystyle \\$

$\displaystyle \text{(ii) } 2^{x+3} = 4^{x-1}$

$\displaystyle 2^{x+3} = 4^{x-1}$

$\displaystyle 2^{x+3} = 2^{2(x-1)}$

$\displaystyle \Rightarrow x+3 = 2(x-1)$

$\displaystyle \Rightarrow x = 5$

$\displaystyle \\$

$\displaystyle \text{(iii) } 2^{2x+1} = 17. 2^x - 2^3$

$\displaystyle 2^{2x+1} = 17. 2^x - 2^3$

$\displaystyle 2.2^{2x} - 17.2^x +8 = 0$

Let $\displaystyle 2^x = k$

$\displaystyle \Rightarrow 2k^2-17k+8=0$

$\displaystyle \Rightarrow (k-8)(2k-1)=0 \Rightarrow k = 8 or k = \frac{1}{2}$

Therefore if $\displaystyle 2^x = 8 \Rightarrow 2^x = 2^3 \Rightarrow then x = 3$

If $\displaystyle 2^x = \frac{1}{2} \Rightarrow 2^x = 2^{-1} \Rightarrow then x = -1$

$\displaystyle \\$

$\displaystyle \text{(iv) } 5^{2x+1} = 6.5^x - 1$

$\displaystyle 5^{2x+1} = 6.5^x - 1$

$\displaystyle 5. 5^{2x} - 6.5^x+1 = 0$

Let $\displaystyle 5^x = k$, Therefore

$\displaystyle 5k^2-6k+1=0$

$\displaystyle (5k-1)(k-1)=0 \Rightarrow k = \frac{1}{5} or k= 1$

Therefore if $\displaystyle 5^x = 1 \Rightarrow 5^x = 5^0 \Rightarrow then x = 0$

If $\displaystyle 5^x = \frac{1}{5} \Rightarrow 5^x = 5^{-1} \Rightarrow then x = -1$

$\displaystyle \\$

$\displaystyle \text{(v) } 2^{2x} - 2^{x+3}+2^4 = 0$

$\displaystyle 2^{2x} - 2^{x+3}+2^4 = 0$

$\displaystyle 2^{2x} - 8.2^x+16 = 0$

Let $\displaystyle 2^x = k$

$\displaystyle k^2 -8k+16 = 0$

$\displaystyle (k-4)(k-4) = 0 \Rightarrow k = 4$

Therefore if $\displaystyle 2^x = 4 \Rightarrow 2^x = 2^2 \Rightarrow then x = 2$

$\displaystyle \\$

$\displaystyle \text{(vi) } 3^{2x+4} + 1 = 2. 3^{x+2}$

$\displaystyle 3^{2x+4} + 1 = 2. 3^{x+2}$

$\displaystyle 81.3^{2x} - 18 3^x +1 = 0$

$\displaystyle (9k-1)(9k-1) = 0 \Rightarrow k = \frac{1}{9}$

If $\displaystyle 3^x = \frac{1}{9} \Rightarrow 3^x = 3^{-2} \Rightarrow then x = -2$

$\displaystyle \\$

Question 10: Given $\displaystyle 4725 = 3^a 5^b 7^c$, find (i) the integral value of $\displaystyle a, b, and c$ (ii) the value of $\displaystyle 2^{-a}3^b7^c$

$\displaystyle 4725 = 3^a 5^b 7^c$
$\displaystyle 5^2.3^3.7^1 = 3^a 5^b 7^c$
$\displaystyle \Rightarrow a=3, b=2, c=1$
Therefore $\displaystyle 2^{-a}3^b7^c = 2^{-3}.3^2.7^1 = \frac{63}{8}$
$\displaystyle \\$