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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

$= \$ $x^0$ $= 1$

$\\$

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

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

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

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

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

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

$= \ 1$

$\\$

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

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

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

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

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

$= \ 1$

$\\$

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

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

$= \ 1$

$\\$

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

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

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

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

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

$= \ 1$

$\\$

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

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

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

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

$= \$ $abc$

$\\$

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

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

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

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

$\\$

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

(i)   $\sqrt{x^{-2}y^3}$

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

$\\$

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

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

$\\$

(iii)   $(\sqrt{x^{-3}})^5$

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

$\\$

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

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

$\\$

(v)   $\sqrt[3]{xy^2} \div x^2y$

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

$\\$

(vi)   $\sqrt[4]{\sqrt[3]{x^2}}$

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

$\\$

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

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

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

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

$\\$

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

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

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

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

$\\$

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

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

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

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

$= 1$

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

$=$  $\frac{x^a+x^b}{x^a+x^b}$

$= 1$

$\\$

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

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

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

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

$= 1$

$\\$

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

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

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

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

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

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

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

$=$ $x^0$

$\\$

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

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

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

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

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

$\\$

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

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

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

$= x^0 =1$

$\\$

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

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

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

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

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

$= x^1 = x$

$\\$

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

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

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

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

$= 1$

$\\$

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

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

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

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

$= 1$

$\\$

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

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

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

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

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

$\\$

Question 4:

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

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

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

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

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

$=$ $x^0.y^0$

$= 1$

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

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

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

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

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

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

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

Therefore RHS = LHS. Hence proved.

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

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

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

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

$=$ $x^0.y^0$

$= 1$

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Question 5:

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

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

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

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

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

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

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

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

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

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

Hence $x+y+2 = 0$

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

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

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

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

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

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Question 7:

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

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

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

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

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

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

Hence Proved.

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

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

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

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

$\Rightarrow 9^x = 3$

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

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

$\\$

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

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

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

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

Similarly, $10^{1/y}= 0.0001$

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

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

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

$\\$

Question 8:

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

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

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

Now, $b^2 = ac$

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

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

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

$\\$

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

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

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

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

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

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

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

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

$\\$

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

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

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

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

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

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

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

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Question 9: Solve the following equations:

(i)   $2^{x-5} = 256$

$2^{x-5} = 256$

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

$\Rightarrow x-5 = 8$

$\Rightarrow x = 13$

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(ii)    $2^{x+3} = 4^{x-1}$

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

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

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

$\Rightarrow x = 5$

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(iii)   $2^{2x+1} = 17. 2^x - 2^3$

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

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

Let $2^x = k$

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

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

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

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

$\\$

(iv)   $5^{2x+1} = 6.5^x - 1$

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

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

Let $5^x = k$, Therefore

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

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

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

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

$\\$

(v)   $2^{2x} - 2^{x+3}+2^4 = 0$

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

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

Let $2^x = k$

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

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

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

$\\$

(vi)   $3^{2x+4} + 1 = 2. 3^{x+2}$

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

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

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

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

$\\$

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

$4725 = 3^a 5^b 7^c$

$5^2.3^3.7^1 = 3^a 5^b 7^c$

$\Rightarrow a=3, b=2, c=1$

Therefore $2^{-a}3^b7^c = 2^{-3}.3^2.7^1 =$ $\frac{63}{8}$

$\\$