\displaystyle \textbf{Question 1: }\text{Assuming that }x\text{ is a positive real number and }a,b,c\text{ are rational}
\displaystyle \text{numbers, show that:}
\displaystyle \text{(i) }\left(\frac{x^b}{x^c}\right)^a\left(\frac{x^c}{x^a}\right)^b\left(\frac{x^a}{x^b}\right)^c=1
\displaystyle \text{(ii) }\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{ac}}=1
\displaystyle \text{(iii) }\left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2}\left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2}  \qquad \left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2}=1
\displaystyle \text{(iv) }\left(\frac{x^a}{x^b}\right)^{a+b}\left(\frac{x^b}{x^c}\right)^{b+c}\left(\frac{x^c}{x^a}\right)^{c+a}=1
\displaystyle \text{(v) }\left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}=1
\displaystyle \text{(vi) }\left(\frac{x^a}{x^{-b}}\right)^{a^2-ab+b^2}\left(\frac{x^b}{x^{-c}}\right)^{b^2-bc+c^2}   \qquad \left(\frac{x^c}{x^{-a}}\right)^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}
\displaystyle \text{(vii) }\left(\frac{x^{a(b-c)}}{x^{b(a-c)}}\right)\div\left(\frac{x^b}{x^a}\right)^c=1
\displaystyle \text{(viii) }\frac{(x^{a+b})^2(x^{b+c})^2(x^{c+a})^2}{(x^ax^bx^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) If }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) }\left(a^{-1}+b^{-1}\right)^{-1}=\frac{ab}{a+b}

\displaystyle \text{(i) }\left(\frac{x^b}{x^c}\right)^a\left(\frac{x^c}{x^a}\right)^b\left(\frac{x^a}{x^b}\right)^c=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^b}{x^c}\right)^a\left(\frac{x^c}{x^a}\right)^b\left(\frac{x^a}{x^b}\right)^c
\displaystyle =\left(x^{b-c}\right)^a\left(x^{c-a}\right)^b\left(x^{a-b}\right)^c
\displaystyle =x^{ab-ac}\times x^{bc-ab}\times x^{ac-bc}
\displaystyle =x^{ab-ac+bc-ab+ac-bc}
\displaystyle =x^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(ii) }\left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{ac}}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^a}{x^b}\right)^{\frac{1}{ab}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{bc}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{ac}}
\displaystyle =\left(x^{a-b}\right)^{\frac{1}{ab}}\left(x^{b-c}\right)^{\frac{1}{bc}}\left(x^{c-a}\right)^{\frac{1}{ac}}
\displaystyle =x^{\frac{a-b}{ab}}\times x^{\frac{b-c}{bc}}\times x^{\frac{c-a}{ac}}
\displaystyle =x^{\frac{1}{b}-\frac{1}{a}}\times x^{\frac{1}{c}-\frac{1}{b}}\times x^{\frac{1}{a}-\frac{1}{c}}
\displaystyle =x^{\left(\frac{1}{b}-\frac{1}{a}\right)+\left(\frac{1}{c}-\frac{1}{b}\right)+\left(\frac{1}{a}-\frac{1}{c}\right)}
\displaystyle =x^0
\displaystyle =1

\displaystyle \text{(iii) }\left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2}\left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2}\left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^a}{x^b}\right)^{a^2+ab+b^2}\left(\frac{x^b}{x^c}\right)^{b^2+bc+c^2}\left(\frac{x^c}{x^a}\right)^{c^2+ca+a^2}
\displaystyle =\left(x^{a-b}\right)^{a^2+ab+b^2}\left(x^{b-c}\right)^{b^2+bc+c^2}\left(x^{c-a}\right)^{c^2+ca+a^2}
\displaystyle =x^{(a-b)(a^2+ab+b^2)}\times x^{(b-c)(b^2+bc+c^2)}\times x^{(c-a)(c^2+ca+a^2)}
\displaystyle =x^{a^3-b^3}\times x^{b^3-c^3}\times x^{c^3-a^3}
\displaystyle =x^{a^3-b^3+b^3-c^3+c^3-a^3}
\displaystyle =x^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(iv) }\left(\frac{x^a}{x^b}\right)^{a+b}\left(\frac{x^b}{x^c}\right)^{b+c}\left(\frac{x^c}{x^a}\right)^{c+a}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^a}{x^b}\right)^{a+b}\left(\frac{x^b}{x^c}\right)^{b+c}\left(\frac{x^c}{x^a}\right)^{c+a}
\displaystyle =\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}
\displaystyle =x^{(a-b)(a+b)}\times x^{(b-c)(b+c)}\times x^{(c-a)(c+a)}
\displaystyle =x^{a^2-b^2}\times x^{b^2-c^2}\times x^{c^2-a^2}
\displaystyle =x^{a^2-b^2+b^2-c^2+c^2-a^2}
\displaystyle =x^0
\displaystyle =1

\displaystyle \text{(v) }\left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^a}{x^b}\right)^{a+b-c}\left(\frac{x^b}{x^c}\right)^{b+c-a}\left(\frac{x^c}{x^a}\right)^{c+a-b}
\displaystyle =\left(x^{a-b}\right)^{a+b-c}\left(x^{b-c}\right)^{b+c-a}\left(x^{c-a}\right)^{c+a-b}
\displaystyle =x^{(a-b)(a+b-c)}\times x^{(b-c)(b+c-a)}\times x^{(c-a)(c+a-b)}
\displaystyle =x^{(a-b)(a+b)-(a-b)c}\times x^{(b-c)(b+c)-(b-c)a}\times x^{(c-a)(c+a)-(c-a)b}
\displaystyle =x^{a^2-b^2-ac+bc}\times x^{b^2-c^2-ab+ac}\times x^{c^2-a^2-bc+ab}
\displaystyle =x^{a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab}
\displaystyle =x^0
\displaystyle =1

\displaystyle \text{(vi) }\left(\frac{x^a}{x^{-b}}\right)^{a^2-ab+b^2}\left(\frac{x^b}{x^{-c}}\right)^{b^2-bc+c^2}\left(\frac{x^c}{x^{-a}}\right)^{c^2-ca+a^2}=x^{2(a^3+b^3+c^3)}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^a}{x^{-b}}\right)^{a^2-ab+b^2}\left(\frac{x^b}{x^{-c}}\right)^{b^2-bc+c^2}\left(\frac{x^c}{x^{-a}}\right)^{c^2-ca+a^2}
\displaystyle =\left(x^{a+b}\right)^{a^2-ab+b^2}\left(x^{b+c}\right)^{b^2-bc+c^2}\left(x^{c+a}\right)^{c^2-ca+a^2}
\displaystyle =x^{(a+b)(a^2-ab+b^2)}\times x^{(b+c)(b^2-bc+c^2)}\times x^{(c+a)(c^2-ca+a^2)}
\displaystyle =x^{a^3+b^3}\times x^{b^3+c^3}\times x^{c^3+a^3}
\displaystyle =x^{2(a^3+b^3+c^3)}
\displaystyle \\

\displaystyle \text{(vii) }\left(\frac{x^{a(b-c)}}{x^{b(a-c)}}\right)\div\left(\frac{x^b}{x^a}\right)^c=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^{a(b-c)}}{x^{b(a-c)}}\right)\div\left(\frac{x^b}{x^a}\right)^c
\displaystyle =\frac{x^{ab-ac}}{x^{ab-bc}}\div\left(x^{b-a}\right)^c
\displaystyle =x^{ab-ac-ab+bc}\div x^{bc-ac}
\displaystyle =x^{-ac+bc}\times x^{-bc+ac}
\displaystyle =x^{-ac+bc-bc+ac}
\displaystyle =x^0
\displaystyle =1

\displaystyle \text{(viii) }\frac{(x^{a+b})^2(x^{b+c})^2(x^{c+a})^2}{(x^ax^bx^c)^4}=1
\displaystyle \text{Answer:}
\displaystyle \frac{(x^{a+b})^2(x^{b+c})^2(x^{c+a})^2}{(x^ax^bx^c)^4}
\displaystyle =\frac{x^{2(a+b)}\times x^{2(b+c)}\times x^{2(c+a)}}{x^{4a}\times x^{4b}\times 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 \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{Answer:}
\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+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 \text{Answer:}
\displaystyle \frac{a^{-1}}{a^{-1}+b^{-1}}+\frac{a^{-1}}{a^{-1}-b^{-1}}
\displaystyle =\frac{\frac1a}{\frac1a+\frac1b}+\frac{\frac1a}{\frac1a-\frac1b}
\displaystyle =\frac{\frac1a}{\frac{a+b}{ab}}+\frac{\frac1a}{\frac{b-a}{ab}}
\displaystyle =\frac1a\times\frac{ab}{a+b}+\frac1a\times\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-ab+ab+b^2}{b^2-a^2}
\displaystyle =\frac{2b^2}{b^2-a^2}

\displaystyle \text{(xi) If }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{Answer:}
\displaystyle \frac{1}{1+a+b^{-1}}+\frac{1}{1+b+c^{-1}}+\frac{1}{1+c+a^{-1}}
\displaystyle =\frac{1}{1+a+\frac1b}+\frac{1}{1+b+\frac1c}+\frac{1}{1+c+\frac1a}
\displaystyle \text{Given, }abc=1\Rightarrow c=\frac1{ab},\quad \frac1c=ab
\displaystyle =\frac{1}{1+a+\frac1b}+\frac{1}{1+b+ab}+\frac{1}{1+\frac1{ab}+\frac1a}
\displaystyle =\frac{b}{b+ab+1}+\frac{1}{b+ab+1}+\frac{ab}{ab+1+b}
\displaystyle =\frac{b+1+ab}{b+ab+1}
\displaystyle =1

\displaystyle \text{(xii) }\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1
\displaystyle \text{Answer:}
\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 \text{Answer:}
\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}{ca}}
\displaystyle =\frac{a+b+c}{\frac{a+b+c}{abc}}
\displaystyle =(a+b+c)\times\frac{abc}{a+b+c}
\displaystyle =abc

\displaystyle \text{(xiv) }\left(a^{-1}+b^{-1}\right)^{-1}=\frac{ab}{a+b}
\displaystyle \text{Answer:}
\displaystyle \left(a^{-1}+b^{-1}\right)^{-1}
\displaystyle =\left(\frac{1}{a}+\frac{1}{b}\right)^{-1}
\displaystyle =\left(\frac{a+b}{ab}\right)^{-1}
\displaystyle =\frac{ab}{a+b}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Assuming that }x,y,z\text{ are positive real numbers, simplify the following:}
\displaystyle \text{(i) }\sqrt{x^{-2}y^3}\qquad \text{(ii) }\left(x^{-\frac23}\right)^2\left(y^{-\frac12}\right)^2
\displaystyle \text{(iii) }\left(\sqrt{x^{-3}}\right)^5
\displaystyle \text{(iv) }\left(\sqrt{x}\right)^{-\frac23}\sqrt{y^4}\div\sqrt{xy^{-\frac12}}
\displaystyle \text{(v) }\sqrt[3]{xy^2}\div x^2y\qquad \text{(vi) }\sqrt[4]{\sqrt[3]{x^2}}
\displaystyle \text{(vii) }\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c} \left(\frac{x^{c+a}}{x^b}\right)^{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{Answer:}
\displaystyle \text{(i) }\sqrt{x^{-2}y^3}
\displaystyle =\sqrt{\frac{y^3}{x^2}}
\displaystyle =\left(\frac{y^3}{x^2}\right)^{\frac12}
\displaystyle =\frac{y^{\frac32}}{x}
\displaystyle \\

\displaystyle \text{(ii) }\left(x^{-\frac23}\right)^2\left(y^{-\frac12}\right)^2
\displaystyle =x^{-\frac43}y^{-1}
\displaystyle =\frac{1}{x^{\frac43}y}
\displaystyle \\

\displaystyle \text{(iii) }\left(\sqrt{x^{-3}}\right)^5
\displaystyle =\left(x^{-\frac32}\right)^5
\displaystyle =x^{-\frac{15}{2}}
\displaystyle =\frac{1}{x^{\frac{15}{2}}}
\displaystyle \\

\displaystyle \text{(iv) }\left(\sqrt{x}\right)^{-\frac23}\sqrt{y^4}\div\sqrt{xy^{-\frac12}}
\displaystyle =\frac{\left(x^{\frac12}\right)^{-\frac23}\left(y^4\right)^{\frac12}}{\left(xy^{-\frac12}\right)^{\frac12}}
\displaystyle =\frac{x^{-\frac13}y^2}{x^{\frac12}y^{-\frac14}}
\displaystyle =x^{-\frac13-\frac12}y^{2+\frac14}
\displaystyle =x^{-\frac56}y^{\frac94}
\displaystyle =\frac{y^{\frac94}}{x^{\frac56}}
\displaystyle \\

\displaystyle \text{(v) }\sqrt[3]{xy^2}\div x^2y
\displaystyle =\frac{(xy^2)^{\frac13}}{x^2y}
\displaystyle =\frac{x^{\frac13}y^{\frac23}}{x^2y}
\displaystyle =x^{-\frac53}y^{-\frac13}
\displaystyle =\frac{1}{x^{\frac53}y^{\frac13}}
\displaystyle \\

\displaystyle \text{(vi) }\sqrt[4]{\sqrt[3]{x^2}}
\displaystyle =\left[\left(x^2\right)^{\frac13}\right]^{\frac14}
\displaystyle =\left(x^{\frac23}\right)^{\frac14}
\displaystyle =x^{\frac16}
\displaystyle \\

\displaystyle \text{(vii) }\left(\frac{x^{a+b}}{x^c}\right)^{a-b}\left(\frac{x^{b+c}}{x^a}\right)^{b-c}\left(\frac{x^{c+a}}{x^b}\right)^{c-a}
\displaystyle =x^{(a+b-c)(a-b)}x^{(b+c-a)(b-c)}x^{(c+a-b)(c-a)}
\displaystyle =x^{a^2-b^2-ac+bc}x^{b^2-c^2-ab+ac}x^{c^2-a^2-bc+ab}
\displaystyle =x^{a^2-b^2-ac+bc+b^2-c^2-ab+ac+c^2-a^2-bc+ab}
\displaystyle =x^0
\displaystyle =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 =\left(\frac{x^l}{x^m}\right)^{\frac{1}{lm}}\left(\frac{x^m}{x^n}\right)^{\frac{1}{mn}}\left(\frac{x^n}{x^l}\right)^{\frac{1}{nl}}
\displaystyle =x^{\frac{l-m}{lm}}x^{\frac{m-n}{mn}}x^{\frac{n-l}{nl}}
\displaystyle =x^{\left(\frac1m-\frac1l\right)+\left(\frac1n-\frac1m\right)+\left(\frac1l-\frac1n\right)}
\displaystyle =x^0
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Assuming that }x,y,z\text{ 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) }\left(\frac{x^{-1}y^2}{x^3y^{-2}}\right)^{\frac13}\div\left(\frac{x^6y^{-3}}{x^{-2}y^3}\right)^{\frac12}=x^ay^b,
\displaystyle \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) }\left\{\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right)\div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right\}^{a+b}=1
\displaystyle \text{(v) }\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}  \left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1
\displaystyle \text{(vi) }\left(\frac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}  \left(\frac{x^{c^2+a^2}}{x^{ac}}\right)^{a+c}=x^{2(a^3+b^3+c^3)}
\displaystyle \text{(vii) }\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}=1
\displaystyle \text{(viii) }\left\{\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right\}^{\frac{a}{a+1}}=x
\displaystyle \text{(ix) }\left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}  \left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}=1
\displaystyle \text{(x) }\left(\frac{3^a}{3^b}\right)^{a+b}\left(\frac{3^b}{3^c}\right)^{b+c}\left(\frac{3^c}{3^a}\right)^{c+a}=1
\displaystyle \text{(xi) }\frac{\left(a+\frac{1}{b}\right)^m\left(a-\frac{1}{b}\right)^n}{\left(b+\frac{1}{a}\right)^m\left(b-\frac{1}{a}\right)^n}=\left(\frac{a}{b}\right)^{m+n}

\displaystyle \text{(i) }\sqrt{x^{-1}y}\times\sqrt{y^{-1}z}\times\sqrt{z^{-1}x}=1
\displaystyle \text{Answer:}
\displaystyle \sqrt{x^{-1}y}\times\sqrt{y^{-1}z}\times\sqrt{z^{-1}x}
\displaystyle =\left(\frac{y}{x}\right)^{\frac12}\times\left(\frac{z}{y}\right)^{\frac12}\times\left(\frac{x}{z}\right)^{\frac12}
\displaystyle =\left(\frac{y}{x}\times\frac{z}{y}\times\frac{x}{z}\right)^{\frac12}
\displaystyle =1^{\frac12}
\displaystyle =1
\displaystyle \\

\displaystyle \text{(ii) If }\left(\frac{x^{-1}y^2}{x^3y^{-2}}\right)^{\frac13}\div\left(\frac{x^6y^{-3}}{x^{-2}y^3}\right)^{\frac12}=x^ay^b,
\displaystyle \text{prove that }a+b=-1,\text{ where }x\text{ and }y\text{ are distinct positive variables.}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^{-1}y^2}{x^3y^{-2}}\right)^{\frac13}\div\left(\frac{x^6y^{-3}}{x^{-2}y^3}\right)^{\frac12}=x^ay^b
\displaystyle \Rightarrow \left(x^{-4}y^4\right)^{\frac13}\div\left(x^8y^{-6}\right)^{\frac12}=x^ay^b
\displaystyle \Rightarrow \frac{x^{-\frac43}y^{\frac43}}{x^4y^{-3}}=x^ay^b
\displaystyle \Rightarrow x^{-\frac43-4}y^{\frac43+3}=x^ay^b
\displaystyle \Rightarrow x^{-\frac{16}{3}}y^{\frac{13}{3}}=x^ay^b
\displaystyle \text{Comparing the exponents of }x\text{ and }y,
\displaystyle a=-\frac{16}{3}\quad\text{and}\quad b=\frac{13}{3}
\displaystyle \therefore a+b=-\frac{16}{3}+\frac{13}{3}
\displaystyle =-\frac{3}{3}
\displaystyle =-1
\displaystyle \\

\displaystyle \text{(iii) }\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1
\displaystyle \text{Answer:}
\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) }\left\{\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right)\div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right\}^{a+b}=1
\displaystyle \text{Answer:}
\displaystyle \left\{\left(\frac{x^{a(a-b)}}{x^{a(a+b)}}\right)\div\left(\frac{x^{b(b-a)}}{x^{b(b+a)}}\right)\right\}^{a+b}
\displaystyle =\left\{x^{a(a-b)-a(a+b)}\div x^{b(b-a)-b(b+a)}\right\}^{a+b}
\displaystyle =\left\{x^{-2ab}\div x^{-2ab}\right\}^{a+b}
\displaystyle =1^{a+b}
\displaystyle =1
\displaystyle \\

\displaystyle \text{(v) }\left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}  \left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}=1
\displaystyle \text{Answer:}
\displaystyle \left(x^{\frac{1}{a-b}}\right)^{\frac{1}{a-c}}\left(x^{\frac{1}{b-c}}\right)^{\frac{1}{b-a}}\left(x^{\frac{1}{c-a}}\right)^{\frac{1}{c-b}}
\displaystyle =x^{\frac{1}{(a-b)(a-c)}+\frac{1}{(b-c)(b-a)}+\frac{1}{(c-a)(c-b)}}
\displaystyle =x^{\frac{b-c-(a-c)+(a-b)}{(a-b)(a-c)(b-c)}}
\displaystyle =x^{\frac{b-c-a+c+a-b}{(a-b)(a-c)(b-c)}}
\displaystyle =x^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(vi) }\left(\frac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}  \left(\frac{x^{c^2+a^2}}{x^{ac}}\right)^{a+c}=x^{2(a^3+b^3+c^3)}
\displaystyle \text{Answer:}
\displaystyle \left(\frac{x^{a^2+b^2}}{x^{ab}}\right)^{a+b}\left(\frac{x^{b^2+c^2}}{x^{bc}}\right)^{b+c}\left(\frac{x^{c^2+a^2}}{x^{ac}}\right)^{a+c}
\displaystyle =x^{(a^2+b^2-ab)(a+b)}\times x^{(b^2+c^2-bc)(b+c)}\times x^{(c^2+a^2-ac)(a+c)}
\displaystyle =x^{a^3+b^3}\times x^{b^3+c^3}\times x^{c^3+a^3}
\displaystyle =x^{2(a^3+b^3+c^3)}

\displaystyle \text{(vii) }\left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}=1
\displaystyle \text{Answer:}
\displaystyle \left(x^{a-b}\right)^{a+b}\left(x^{b-c}\right)^{b+c}\left(x^{c-a}\right)^{c+a}
\displaystyle =x^{a^2-b^2}\times x^{b^2-c^2}\times x^{c^2-a^2}
\displaystyle =x^{a^2-b^2+b^2-c^2+c^2-a^2}
\displaystyle =x^0
\displaystyle =1

\displaystyle \text{(viii) }\left\{\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right\}^{\frac{a}{a+1}}=x
\displaystyle \text{Answer:}
\displaystyle \left\{\left(x^{a-a^{-1}}\right)^{\frac{1}{a-1}}\right\}^{\frac{a}{a+1}}
\displaystyle =\left\{\left(x^{\frac{a^2-1}{a}}\right)^{\frac{1}{a-1}}\right\}^{\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
\displaystyle =x
\displaystyle \\

\displaystyle \text{(ix) }\left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}  \left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{a^{x+1}}{a^{y+1}}\right)^{x+y}\left(\frac{a^{y+2}}{a^{z+2}}\right)^{y+z}\left(\frac{a^{z+3}}{a^{x+3}}\right)^{z+x}
\displaystyle =\left(a^{x-y}\right)^{x+y}\times\left(a^{y-z}\right)^{y+z}\times\left(a^{z-x}\right)^{z+x}
\displaystyle =a^{x^2-y^2}\times a^{y^2-z^2}\times a^{z^2-x^2}
\displaystyle =a^{x^2-y^2+y^2-z^2+z^2-x^2}
\displaystyle =a^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(x) }\left(\frac{3^a}{3^b}\right)^{a+b}\left(\frac{3^b}{3^c}\right)^{b+c}\left(\frac{3^c}{3^a}\right)^{c+a}=1
\displaystyle \text{Answer:}
\displaystyle \left(\frac{3^a}{3^b}\right)^{a+b}\left(\frac{3^b}{3^c}\right)^{b+c}\left(\frac{3^c}{3^a}\right)^{c+a}
\displaystyle =\left(3^{a-b}\right)^{a+b}\times\left(3^{b-c}\right)^{b+c}\times\left(3^{c-a}\right)^{c+a}
\displaystyle =3^{a^2-b^2}\times3^{b^2-c^2}\times3^{c^2-a^2}
\displaystyle =3^{a^2-b^2+b^2-c^2+c^2-a^2}
\displaystyle =3^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(xi) }\frac{\left(a+\frac1b\right)^m\left(a-\frac1b\right)^n}{\left(b+\frac1a\right)^m\left(b-\frac1a\right)^n}=\left(\frac ab\right)^{m+n}
\displaystyle \text{Answer:}
\displaystyle \frac{\left(a+\frac1b\right)^m\left(a-\frac1b\right)^n}{\left(b+\frac1a\right)^m\left(b-\frac1a\right)^n}
\displaystyle =\frac{\left(\frac{ab+1}{b}\right)^m\left(\frac{ab-1}{b}\right)^n}{\left(\frac{ab+1}{a}\right)^m\left(\frac{ab-1}{a}\right)^n}
\displaystyle =\frac{(ab+1)^m(ab-1)^n}{b^{m+n}}\times\frac{a^{m+n}}{(ab+1)^m(ab-1)^n}
\displaystyle =\frac{a^{m+n}}{b^{m+n}}
\displaystyle =\left(\frac ab\right)^{m+n}
\displaystyle \\

\displaystyle \textbf{Question 4:}
\displaystyle \text{(i) If }a=x^{m+n}y^l,\ b=x^{n+l}y^m\text{ and }c=x^{l+m}y^n,\text{ prove that}
\displaystyle a^{m-n}b^{n-l}c^{l-m}=1
\displaystyle \text{(ii) If }x=a^{m+n},\ y=a^{n+l}\text{ and }z=a^{l+m},\text{ prove that}
\displaystyle x^my^nz^l=x^ny^lz^m
\displaystyle \text{(iii) If }a=xy^{p-1},\ b=xy^{q-1}\text{ and }c=xy^{r-1},\text{ prove that}
\displaystyle a^{q-r}b^{r-p}c^{p-q}=1
\displaystyle \text{Answer:}
\displaystyle \text{(i) }a^{m-n}b^{n-l}c^{l-m}
\displaystyle =\left(x^{m+n}y^l\right)^{m-n}\times\left(x^{n+l}y^m\right)^{n-l}\times\left(x^{l+m}y^n\right)^{l-m}
\displaystyle =x^{(m+n)(m-n)+(n+l)(n-l)+(l+m)(l-m)}
\displaystyle \qquad\times y^{l(m-n)+m(n-l)+n(l-m)}
\displaystyle =x^{m^2-n^2+n^2-l^2+l^2-m^2}
\displaystyle \qquad\times y^{lm-ln+mn-ml+nl-nm}
\displaystyle =x^0y^0
\displaystyle =1
\displaystyle \\

\displaystyle \text{(ii) LHS}=x^my^nz^l
\displaystyle =\left(a^{m+n}\right)^m\left(a^{n+l}\right)^n\left(a^{l+m}\right)^l
\displaystyle =a^{m(m+n)+n(n+l)+l(l+m)}
\displaystyle =a^{m^2+n^2+l^2+mn+nl+lm}
\displaystyle \text{RHS}=x^ny^lz^m
\displaystyle =\left(a^{m+n}\right)^n\left(a^{n+l}\right)^l\left(a^{l+m}\right)^m
\displaystyle =a^{n(m+n)+l(n+l)+m(l+m)}
\displaystyle =a^{m^2+n^2+l^2+mn+nl+lm}
\displaystyle \therefore \text{LHS}=\text{RHS}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{(iii) }a^{q-r}b^{r-p}c^{p-q}
\displaystyle =\left(xy^{p-1}\right)^{q-r}\left(xy^{q-1}\right)^{r-p}\left(xy^{r-1}\right)^{p-q}
\displaystyle =x^{q-r+r-p+p-q}
\displaystyle \qquad\times y^{(p-1)(q-r)+(q-1)(r-p)+(r-1)(p-q)}
\displaystyle =x^0y^{pq-pr-q+r+qr-qp-r+p+rp-rq-p+q}
\displaystyle =x^0y^0
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 5:}
\displaystyle \text{(i) If }a\text{ and }b\text{ are distinct positive primes such that}
\displaystyle \sqrt[3]{a^6b^{-4}}=a^xb^{2y},\text{ find }x\text{ and }y.
\displaystyle \text{(ii) If }a\text{ and }b\text{ are distinct positive primes such that}
\displaystyle (a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^xb^y,\text{ find }x+y+2.
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\sqrt[3]{a^6b^{-4}}=a^xb^{2y}
\displaystyle \Rightarrow \left(a^6b^{-4}\right)^{\frac13}=a^xb^{2y}
\displaystyle \Rightarrow a^2b^{-\frac43}=a^xb^{2y}
\displaystyle \text{Comparing the exponents of }a\text{ and }b,
\displaystyle x=2\quad\text{and}\quad 2y=-\frac43
\displaystyle \therefore x=2\quad\text{and}\quad y=-\frac23
\displaystyle \\

\displaystyle \text{(ii) }(a+b)^{-1}\left(a^{-1}+b^{-1}\right)=a^xb^y
\displaystyle \Rightarrow \frac{1}{a+b}\left(\frac1a+\frac1b\right)=a^xb^y
\displaystyle \Rightarrow \frac{1}{a+b}\left(\frac{a+b}{ab}\right)=a^xb^y
\displaystyle \Rightarrow a^{-1}b^{-1}=a^xb^y
\displaystyle \text{Comparing the exponents of }a\text{ and }b,
\displaystyle x=-1\quad\text{and}\quad y=-1
\displaystyle \therefore x+y+2=-1-1+2
\displaystyle =0
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }2^x\times3^y\times5^z=2160,\text{ find }x,y\text{ and }z.
\displaystyle \text{Then compute the value of }3^x\times2^{-y}\times5^{-z}.
\displaystyle \text{Answer:}
\displaystyle 2160=2\times2\times2\times2\times3\times3\times3\times5
\displaystyle =2^4\times3^3\times5^1
\displaystyle \text{Given, }2^x\times3^y\times5^z=2^4\times3^3\times5^1
\displaystyle \text{Comparing the powers of the distinct primes }2,3\text{ and }5,
\displaystyle x=4,\quad y=3\quad\text{and}\quad z=1
\displaystyle \therefore 3^x\times2^{-y}\times5^{-z}
\displaystyle =3^4\times2^{-3}\times5^{-1}
\displaystyle =81\times\frac18\times\frac15
\displaystyle =\frac{81}{40}
\displaystyle \\

\displaystyle \textbf{Question 7:}
\displaystyle \text{(i) If }x=2^{\frac13}+2^{\frac23},\text{ show that }x^3-6x=6.
\displaystyle \text{(ii) Determine }\left(8x\right)^x,\text{ if }9^{x+2}=240+9^x.
\displaystyle \text{(iii) If }3^{4x}=81^{-1}\text{ and }10^{\frac1y}=0.0001,\text{ find the value of}
\displaystyle 2^{-x+4y}.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Given, }x=2^{\frac13}+2^{\frac23}
\displaystyle \text{We know that }\left(a+b\right)^3=a^3+b^3+3ab(a+b)
\displaystyle \therefore x^3=\left(2^{\frac13}+2^{\frac23}\right)^3
\displaystyle =2+4+3\left(2^{\frac13}\times2^{\frac23}\right)\left(2^{\frac13}+2^{\frac23}\right)
\displaystyle =6+6\left(2^{\frac13}+2^{\frac23}\right)
\displaystyle =6+6x
\displaystyle \therefore x^3-6x=6
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) Given, }9^{x+2}=240+9^x
\displaystyle \Rightarrow 9^{x+2}-9^x=240
\displaystyle \Rightarrow 9^x\left(9^2-1\right)=240
\displaystyle \Rightarrow 9^x\left(81-1\right)=240
\displaystyle \Rightarrow 80\times9^x=240
\displaystyle \Rightarrow 9^x=3
\displaystyle \Rightarrow \left(3^2\right)^x=3^1
\displaystyle \Rightarrow 2x=1
\displaystyle \Rightarrow x=\frac12
\displaystyle \therefore \left(8x\right)^x=\left(8\times\frac12\right)^{\frac12}
\displaystyle =4^{\frac12}
\displaystyle =2
\displaystyle \\

\displaystyle \text{(iii) Given, }3^{4x}=81^{-1}
\displaystyle \Rightarrow 3^{4x}=\left(3^4\right)^{-1}
\displaystyle \Rightarrow 3^{4x}=3^{-4}
\displaystyle \Rightarrow 4x=-4
\displaystyle \Rightarrow x=-1
\displaystyle \text{Also, }10^{\frac1y}=0.0001
\displaystyle \Rightarrow 10^{\frac1y}=10^{-4}
\displaystyle \Rightarrow \frac1y=-4
\displaystyle \Rightarrow y=-\frac14
\displaystyle \therefore 2^{-x+4y}=2^{-(-1)+4\left(-\frac14\right)}
\displaystyle =2^{1-1}
\displaystyle =2^0
\displaystyle =1
\displaystyle \\

\displaystyle \textbf{Question 8:}
\displaystyle \text{(i) If }a^x=b^y=c^z\text{ and }b^2=ac,\text{ show that }y=\frac{2zx}{z+x}.
\displaystyle \text{(ii) If }2^x=3^y=6^{-z},\text{ show that }\frac1x+\frac1y+\frac1z=0.
\displaystyle \text{(iii) If }2^x=3^y=12^z,\text{ show that }\frac1z=\frac1y+\frac2x.
\displaystyle \text{Answer:}
\displaystyle \text{(i) Let }a^x=b^y=c^z=k.
\displaystyle \therefore a=k^{\frac1x},\quad b=k^{\frac1y}\quad\text{and}\quad c=k^{\frac1z}
\displaystyle \text{Given, }b^2=ac
\displaystyle \Rightarrow \left(k^{\frac1y}\right)^2=k^{\frac1x}\times k^{\frac1z}
\displaystyle \Rightarrow k^{\frac2y}=k^{\frac1x+\frac1z}
\displaystyle \Rightarrow \frac2y=\frac1x+\frac1z
\displaystyle \Rightarrow \frac2y=\frac{x+z}{xz}
\displaystyle \therefore y=\frac{2xz}{x+z}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{(ii) Let }2^x=3^y=6^{-z}=k.
\displaystyle \therefore 2=k^{\frac1x},\quad 3=k^{\frac1y}\quad\text{and}\quad 6=k^{-\frac1z}
\displaystyle \text{Since }6=2\times3,
\displaystyle k^{-\frac1z}=k^{\frac1x}\times k^{\frac1y}
\displaystyle \Rightarrow k^{-\frac1z}=k^{\frac1x+\frac1y}
\displaystyle \Rightarrow -\frac1z=\frac1x+\frac1y
\displaystyle \therefore \frac1x+\frac1y+\frac1z=0
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \text{(iii) Let }2^x=3^y=12^z=k.
\displaystyle \therefore 2=k^{\frac1x},\quad 3=k^{\frac1y}\quad\text{and}\quad 12=k^{\frac1z}
\displaystyle \text{Since }12=2^2\times3,
\displaystyle k^{\frac1z}=\left(k^{\frac1x}\right)^2\times k^{\frac1y}
\displaystyle \Rightarrow k^{\frac1z}=k^{\frac2x+\frac1y}
\displaystyle \Rightarrow \frac1z=\frac2x+\frac1y
\displaystyle \therefore \frac1z=\frac1y+\frac2x
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{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\times2^x-2^3
\displaystyle \text{(iv) }5^{2x+1}=6\times5^x-1
\displaystyle \text{(v) }2^{2x}-2^{x+3}+2^4=0
\displaystyle \text{(vi) }3^{2x+4}+1=2\times3^{x+2}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }2^{x-5}=256
\displaystyle \Rightarrow 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 \Rightarrow 2^{x+3}=\left(2^2\right)^{x-1}
\displaystyle \Rightarrow 2^{x+3}=2^{2x-2}
\displaystyle \Rightarrow x+3=2x-2
\displaystyle \Rightarrow x=5
\displaystyle \\

\displaystyle \text{(iii) }2^{2x+1}=17\times2^x-2^3
\displaystyle \Rightarrow 2\times2^{2x}-17\times2^x+8=0
\displaystyle \text{Let }2^x=k.
\displaystyle \therefore 2k^2-17k+8=0
\displaystyle \Rightarrow (k-8)(2k-1)=0
\displaystyle \Rightarrow k=8\quad\text{or}\quad k=\frac12
\displaystyle \text{If }2^x=8,\text{ then }2^x=2^3
\displaystyle \Rightarrow x=3
\displaystyle \text{If }2^x=\frac12,\text{ then }2^x=2^{-1}
\displaystyle \Rightarrow x=-1
\displaystyle \therefore x=3\quad\text{or}\quad x=-1
\displaystyle \\

\displaystyle \text{(iv) }5^{2x+1}=6\times5^x-1
\displaystyle \Rightarrow 5\times5^{2x}-6\times5^x+1=0
\displaystyle \text{Let }5^x=k.
\displaystyle \therefore 5k^2-6k+1=0
\displaystyle \Rightarrow (5k-1)(k-1)=0
\displaystyle \Rightarrow k=\frac15\quad\text{or}\quad k=1
\displaystyle \text{If }5^x=\frac15,\text{ then }5^x=5^{-1}
\displaystyle \Rightarrow x=-1
\displaystyle \text{If }5^x=1,\text{ then }5^x=5^0
\displaystyle \Rightarrow x=0
\displaystyle \therefore x=-1\quad\text{or}\quad x=0
\displaystyle \\

\displaystyle \text{(v) }2^{2x}-2^{x+3}+2^4=0
\displaystyle \Rightarrow 2^{2x}-8\times2^x+16=0
\displaystyle \text{Let }2^x=k.
\displaystyle \therefore k^2-8k+16=0
\displaystyle \Rightarrow (k-4)^2=0
\displaystyle \Rightarrow k=4
\displaystyle \Rightarrow 2^x=4
\displaystyle \Rightarrow 2^x=2^2
\displaystyle \therefore x=2
\displaystyle \\

\displaystyle \text{(vi) }3^{2x+4}+1=2\times3^{x+2}
\displaystyle \Rightarrow 81\times3^{2x}-18\times3^x+1=0
\displaystyle \text{Let }3^x=k.
\displaystyle \therefore 81k^2-18k+1=0
\displaystyle \Rightarrow (9k-1)^2=0
\displaystyle \Rightarrow k=\frac19
\displaystyle \Rightarrow 3^x=\frac19
\displaystyle \Rightarrow 3^x=3^{-2}
\displaystyle \therefore x=-2
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Given }4725=3^a5^b7^c,\text{ find }(i)\text{ the integral values of }a,b\text{ and }c
\displaystyle \text{(ii) the value of }2^{-a}3^b7^c.
\displaystyle \text{Answer:}
\displaystyle 4725=5\times945
\displaystyle =5\times5\times189
\displaystyle =5^2\times3\times63
\displaystyle =5^2\times3^2\times21
\displaystyle =5^2\times3^3\times7
\displaystyle =3^3\times5^2\times7^1
\displaystyle \text{Comparing the powers of the distinct primes }3,5\text{ and }7,
\displaystyle a=3,\quad b=2\quad\text{and}\quad c=1
\displaystyle \therefore 2^{-a}3^b7^c
\displaystyle =2^{-3}\times3^2\times7
\displaystyle =\frac{1}{8}\times9\times7
\displaystyle =\frac{63}{8}
\displaystyle \\

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