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

Answer:

\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

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(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

Answer:

\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

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(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

Answer:

\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

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(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

Answer:

\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

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(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

Answer:

\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

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(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)}

Answer:

\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)}

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

Answer:

\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

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

Answer:

\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

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(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

Answer:

\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

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

Answer:

\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} 

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(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

Answer:

\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

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

Answer:

\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

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

Answer:

\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

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

Answer:

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

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

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

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

(i)   \sqrt{x^{-2}y^3}

Answer:

\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}

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

Answer:

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

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(iii)   (\sqrt{x^{-3}})^5 

Answer:

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

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

Answer:

(\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}}}

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(v)   \sqrt[3]{xy^2} \div x^2y

Answer:

\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}}}    

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(vi)   \sqrt[4]{\sqrt[3]{x^2}}

Answer:

\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}}

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(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}

Answer:

\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

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

Answer:

\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

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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

Answer:

\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

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(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.

Answer:

\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

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

Answer:

\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

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(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

Answer:

\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

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(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

Answer:

\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

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(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)}

Answer:

\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)}

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

Answer:

(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

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

Answer:

\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

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(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

Answer:

\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

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(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

Answer:

\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

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(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}

Answer:

\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}

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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

Answer:

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

Answer:

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

Answer:

(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

Answer:

\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

Answer:

(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}

Answer:

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

Answer:

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

Answer:

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

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

Answer:

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 

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Question 8:

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

Answer:

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}

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

Answer:

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

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

Answer:

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

Answer:

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}

Answer:

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

Answer:

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

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(iv)   5^{2x+1} = 6.5^x - 1

Answer:

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

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(v)   2^{2x} - 2^{x+3}+2^4 = 0

Answer:

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

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(vi)   3^{2x+4} + 1 = 2. 3^{x+2}

Answer:

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

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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

Answer:

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}

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