[4 marks sum]

Question 14 Marks

Simplify: $\left[\frac{256 a ^{16}}{81 b ^4}\right]^{-\frac{3}{4}}$

Answer

$ {\left[\frac{256 a ^{16}}{81 b ^4}\right]^{-\frac{3}{4}}=\left[\frac{4^4 a ^{16}}{3^4 b ^4}\right]^{-\frac{3}{4}}}$
$ =\frac{4^{4 x -\frac{3}{4}} \cdot a ^{16 x -\frac{3}{4}}}{3^{4 x -\frac{3}{4}} \cdot b ^{4 x -\frac{3}{4}}}$
$ =\frac{4^{-3} \cdot a ^{-12}}{3^{-3} \cdot b ^{-3}}$
$ =\frac{3^3 b ^3}{4^3 a ^{12}} \cdots \cdots \cdots$
$ \left(\because 4^{-3}=\frac{1}{4^3}, \frac{1}{3^{-3}}=3^3, a ^{-12}=\frac{1}{ a ^{12}}, \frac{1}{ b ^{-3}}=\prime b^3\right)$
$ =\frac{27 b ^3}{64 a ^{12}}$
$ =\frac{27}{64} \cdot a ^{-12} b ^3$

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Question 24 Marks

Simplify: $\sqrt[5]{x^{20} y^{-10} z^5} \div \frac{x^3}{y^3}$

Answer

$ \sqrt[5]{x^{20} y^{-10} z^5} \div \frac{x^3}{y^3}$
$ =\left(x^{20} y^{-10} z^5\right)^{\frac{1}{5}} \div \frac{x^3}{y^3}$
$ =x^{20 \times \frac{1}{5}} \cdot y^{-10 \times \frac{1}{5}} \cdot z^{5 \times \frac{1}{5}} \div \frac{x^3}{y^3}$
$ =x^4 \cdot y^{-2} \cdot z^1 \times \frac{y^3}{x^3}$
$ =x^{4-3} \cdot y^{-2+3} \cdot z^1$
$ =x y z$

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Question 34 Marks

Evaluate: $9^0+9^{-1}-9^{-2}+9^{\frac{1}{2}}-9^{-\frac{1}{2}}$

Answer

$ 9^0+9^{-1}-9^{-2}+9^{\frac{1}{2}}-9^{-\frac{1}{2}}$
$ =1+\frac{1}{9}-\frac{1}{9^2}+\left(3^2\right)^{\frac{1}{2}}-\left(3^2\right)^{-\frac{1}{2}}$
$ =1+\frac{1}{9}-\frac{1}{81}+3^{2 \times \frac{1}{2}}-3^{2 \times\left(-\frac{1}{2}\right)}$
$ =1+\frac{1}{9}-\frac{1}{81}+\frac{3}{1}-\frac{1}{3}$
$ =\frac{81+9-1+243-27}{81}$
$=\frac{333-28}{81}$
$ =\frac{305}{81}$
$=3 \frac{62}{81}$

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Question 44 Marks

Simplify: $\left[(64)^{-2}\right]^{-3} \div\left[\left\{(-8)^2\right\}^3\right]^2$

Answer

$ {\left[(64)^{-2}\right]^{-3} \div\left[\left\{(-8)^2\right\}^3\right]^2}$
$ =\left(2^6\right)^{-2 \times-3 \div(-8)^{2 \times 3 \times 2}}$
$ =2^{6 \times(6) \div(-8)^{12}}$
$ =2^{+36} \div(-8)^{12}$
$ =2^{+36} \div\left[(-2)^3\right]^{12}=2^{36} \div(-2)^{36}$
$ =\frac{2^{36}}{(-2)^{36}}=\frac{2^{36}}{2^{36}} \quad(\because 36$ is even $)$
$ =2^{36-36}=2^0=1 \quad\left(\because a ^0=1\right)$

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Question 54 Marks

Find the value of $n$, when:$\frac{ a ^{2 n -3} \times\left( a ^2\right)^{ n +1}}{\left( a ^4\right)^{-3}}=\left( a ^3\right)^3 \div\left( a ^6\right)^{-3}$

Answer

$\frac{a^{2 n-3} \times\left(a^2\right)^{n+1}}{\left(a^4\right)^{-3}}=\left(a^3\right)^3 \div\left(a^6\right)^{-3}$
$ \frac{a^{2 n-3} \times 2^{2 n+2}}{a^{-12}}=a^9 \div a^{-18}$
$ \frac{a^{2 n-3} \times 2^{2 n+2}}{a^{-12}}=\frac{a^9}{a^{-18}}$
$ a^{2 n-3+2 n+2-(-12)=a^9-(-18)}$
$ a^{4 n+11}=a^{27}$
Comparing both sides, we get
$4 n +11=27$
$ \Rightarrow 4 n =27-11$
$ \Rightarrow n =\frac{16}{4}=4$

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Question 64 Marks

Compute: $(-3)^4-(\sqrt[4]{3})^0 \times(-2)^5 \div(64)^{\frac{2}{3}}$

Answer

$ (-3)^4-(\sqrt[4]{3})^0 \times(-2)^5 \div(64)^{\frac{2}{3}}$
$ =(-3 \times-3 \times-3 \times-3)-1 \times-2 \times-2 \times-2 \times-2$
$ \text { Note }:(\sqrt[4]{3})^0=1$
$ =3^4+2^5 \div 2^{6 \times \frac{2}{3}}$
$ =3^4+2^5 \div 2^4=3^4+\frac{2^5}{2^4}$
$ =3^4+2^{5-4}=3^4+2=3 \times 3 \times 3 \times 3+2$
$ =81+2=83$

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Question 74 Marks

Compute: $(243)^{\frac{2}{5}} \div(32)^{-\frac{2}{5}}$

Answer

$ (243)^{\frac{2}{5}} \div(32)^{-\frac{2}{5}}$
$ =(3 \times 3 \times 3 \times 3 \times 3)^{\frac{2}{5}} \div(2 \times 2 \times 2 \times 2 \times 2)^{-\frac{2}{5}}$
$ =\left(3^5\right)^{\frac{2}{5}} \div\left(2^5\right)^{-\frac{2}{5}}$
$ =3^{5 \times \frac{2}{5}} \div 2^{-\frac{2}{5} \times 5}=3^2 \div 2^{-2}$
$ =3^2 \times \frac{1}{2^{-2}}=3^2 \times 2^{+2}$
$ =3 \times 3 \times 2 \times 2=36$

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Question 84 Marks

Prove that : $\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1\ A$

Answer

$\frac{1}{1+x^{a-b}}+\frac{1}{1+x^{b-a}}=1$
$\text{L.H.S} =\frac{1}{1+ x ^{ a - b }}+\frac{1}{1+ x ^{ b - a }}$
$=\frac{1}{ x ^{ a - a }+ x ^{ a - b }}+\frac{1}{ x ^{ b - b }+ x ^{ b - a }}$
$=\frac{1}{ x ^{ a } \cdot x ^{- a }+ x ^{ a } \cdot x ^{- b }}+\frac{1}{ x ^{ b } \cdot x ^{- b }+ x ^{ b } \cdot x ^{- a }}$
$=\frac{1}{ x ^{ a }\left( x ^{- a }+ x ^{- b }\right)}+\frac{1}{ x ^{ b }\left( x ^{- b }+ x ^{- a }\right)}$
$=\frac{1}{\left( x ^{- a }+ x ^{- b }\right)}\left[\frac{1}{ x ^{ a }}+\frac{1}{ x ^{ b }}\right]$
$=\frac{1}{ x ^{- a }+ x ^{- b }}\left[ x ^{- a }+ x ^{- b }\right]=1=\text { R. H.S. }$

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Question 94 Marks

Prove that : $\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}=1$

Answer

$\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}=1$
$\text{L. H.S}. =\left(\frac{x^a}{x^b}\right)^{\frac{1}{a b}}\left(\frac{x^b}{x^c}\right)^{\frac{1}{b c}}\left(\frac{x^c}{x^a}\right)^{\frac{1}{c a}}$
$\left(x^{a-b}\right)^{\frac{1}{a b}}\left(x^{b-c}\right)^{\frac{1}{b c}}\left(x^{c-a}\right)^{\frac{1}{c a}}$
$ =x^{\frac{a-b}{a b}} x^{\frac{b-c}{b c}} x^{\frac{c-a}{c a}} \ldots \ldots \ldots \ldots\left\{\left(x^a\right)^b=x^{a b}\right\}$
$=x^{\frac{ a - b }{ ab }+\frac{ b - c }{ bc }+\frac{ c - a }{ ca }}$
$=x^{\frac{a c-b c+a b-a c+b c-a b}{a b c}}$
$=x^0=1= \text{R. H.S}.\left(\because x^0=1\right)$

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