Surds And Indices — Part 1 MCQs with Explanations

Let (17) 3.5 × (17) x = 17 8 Then, (17) 3.5 + x = 17 8 ∴ 3.5 + x = 8 ⇒ x = (8 - 3.5) ⇒ x = 4.5

$$\eqalign{ & {\text{Given}}\,{\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{b}{a}} \right)^{x - 3}} \cr & \Rightarrow {\left( {\frac{a}{b}} \right)^{x - 1}} = {\left( {\frac{a}{b}} \right)^{ - \left( {x - 3} \right)}} = {\left( {\frac{a}{b}} \right)^{\left( {3 - x} \right)}} \cr & \Rightarrow x - 1 = 3 - x \cr & \Rightarrow 2x = 4 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & {x^z} = {y^2} \Leftrightarrow {10^{\left( {0.48z} \right)}} = {10^{2 \times 0.70}} = {10^{1.40}} \cr & \Rightarrow 0.48z = 1.40 \cr & \Rightarrow z = \frac{{140}}{{48}} = \frac{{35}}{{12}} = 2.9({\text{approx}}) \cr} $$

5 a = 3125 ⇔ 5 a = 5 5 ⇒ a = 5. ∴ 5 ( a - 3) = 5 (5 - 3) = 5 2 = 25

3 x - y = 27 = 3 3 ⇔ x - y = 3 ....(i) 3 x + y = 243 = 3 5 ⇔ x + y = 5 ....(ii) On solving (i) and (ii), we get x = 4.

(256) 0.16 × (256) 0.09 = (256) (0.16 + 0.09) $$\eqalign{ & = {\left( {256} \right)^{0.25}} \cr & = {\left( {256} \right)^{\frac{{25}}{{100}}}} \cr & = {\left( {256} \right)^{\frac{1}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{1}{4}}} \cr & = {4^1} \cr & = 4 \cr} $$

$$\eqalign{ & {\left( {10} \right)^{150}} \div {\left( {10} \right)^{146}} = \frac{{{{10}^{150}}}}{{{{10}^{146}}}} \cr & = {10^{150 - 146}} \cr & = {10^4} \cr & = 10000 \cr} $$

Given exp. = $$ = \frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^a}}} + \frac{{{x^c}}}{{{x^a}}}} \right)}} + $$ $$\frac{1}{{\left( {1 + \frac{{{x^a}}}{{{x^b}}} + \frac{{{x^c}}}{{{x^b}}}} \right)}} + $$ $$\frac{1}{{\left( {1 + \frac{{{x^b}}}{{{x^c}}} + \frac{{{x^a}}}{{{x^c}}}} \right)}}$$ $$ = \frac{{{x^a}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$ $$\frac{{{x^b}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}} + $$ $$\frac{{{x^c}}}{{\left( {{x^a} + {x^b} + {x^c}} \right)}}$$ $$\eqalign{ & = \frac{{ {{x^a} + {x^b} + {x^c}} }}{{ {{x^a} + {x^b} + {x^c}} }} \cr & = 1 \cr} $$

$$\eqalign{ & {\text{Let}}\,{\left( {25} \right)^{7.5}} \times {\left( 5 \right)^{2.5}} \div {\left( {125} \right)^{1.5}} = {5^x} \cr & {\text{Then}},\,\frac{{{{\left( {{5^2}} \right)}^{7.5}} \times {{\left( 5 \right)}^{2.5}}}}{{{{\left( {{5^3}} \right)}^{1.5}}}} = {5^x} \cr & \Rightarrow \frac{{{5^{\left( {2 \times 7.5} \right)}} \times {5^{2.5}}}}{{{5^{\left( {3 \times 1.5} \right)}}}} = {5^x} \cr & \Rightarrow \frac{{{5^{15}} \times {5^{2.5}}}}{{{5^{4.5}}}} = {5^x} \cr & \Rightarrow {5^x} = {5^{\left( {15 + 2.5 - 4.5} \right)}} \cr & \Rightarrow {5^x} = {5^{13}} \cr & \therefore x = 13 \cr} $$

$$\eqalign{ & {\left( {0.04} \right)^{ - 1.5}} = {\left( {\frac{4}{{100}}} \right)^{ - 1.5}} \cr & = {\left( {\frac{1}{{25}}} \right)^{ - \left( {3/2} \right)}} \cr & = {\left( {25} \right)^{\left( {3/2} \right)}} \cr & = {\left( {{5^2}} \right)^{\left( {3/2} \right)}} \cr & = {\left( 5 \right)^{2 \times \left( {3/2} \right)}} \cr & = {5^3} \cr & = 125 \cr} $$

$$\eqalign{ & {\text{Given}}\,{\text{Expression}} \cr & = \frac{{{{\left( {243} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{9^n} \times {3^{n - 1}}}} \cr & = \frac{{{{\left( {{3^5}} \right)}^{n/5}} \times {3^{2n + 1}}}}{{{{\left( {{3^2}} \right)}^n} \times {3^{n - 1}}}} \cr & = \frac{{\left( {{3^{5 \times \left( {n/5} \right)}} \times {3^{2n + 1}}} \right)}}{{\left( {{3^{2n}} \times {3^{n - 1}}} \right)}} \cr & = \frac{{{3^n} \times {3^{2n + 1}}}}{{{3^{2n}} \times {3^{n - 1}}}} \cr & = \frac{{{3^{\left( {n + 2n + 1} \right)}}}}{{{3^{\left( {2n + n - 1} \right)}}}} \cr & = \frac{{{3^{3n + 1}}}}{{{3^{3n - 1}}}} \cr & = {3^{\left( {3n + 1 - 3n + 1} \right)}} \cr & = {3^2} \cr & = 9 \cr} $$

$$\eqalign{ & \frac{1}{{1 + {a^{\left( {n - m} \right)}}}} + \frac{1}{{1 + {a^{\left( {m - n} \right)}}}} \cr & = \frac{1}{{\left( {1 + \frac{{{a^n}}}{{{a^m}}}} \right)}} + \frac{1}{{\left( {1 + \frac{{{a^m}}}{{{a^n}}}} \right)}} \cr & = \frac{{{a^m}}}{{\left( {{a^m} + {a^n}} \right)}} + \frac{{{a^n}}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = \frac{{\left( {{a^m} + {a^n}} \right)}}{{\left( {{a^m} + {a^n}} \right)}} \cr & = 1 \cr} $$

We know that 11 2 = 121. Putting m = 11 and n = 2, we get: ( m - 1) n + 1 = (11 - 1) (2 + 1) = 10 3 = 1000

$$\eqalign{ & {\text{Given Exp}}{\text{.}} \cr & = {x^{\left( {b - c} \right)\left( {b + c - a} \right)}}.{x^{\left( {c - a} \right)\left( {c + a - b} \right)}}.{x^{\left( {a - b} \right)\left( {a + b - c} \right)}} \cr} $$ $$ = {x^{\left( {b - c} \right)\left( {b + c} \right) - a\left( {b - c} \right)}}.$$ $${x^{\left( {c - a} \right)\left( {c + a} \right) - b\left( {c - a} \right)}}.$$ $${x^{\left( {a - b} \right)\left( {a + b} \right) - c\left( {a - b} \right)}}$$ $$\eqalign{ & = {x^{\left( {{b^2} - {c^2} + {c^2} - {a^2} + {a^2} - {b^2}} \right)}}.{x^{ - a\left( {b - c} \right) - b\left( {c - a} \right) - c\left( {a - b} \right)}} \cr & = {{x^0} . {x^\left( -ab + ac - bc + ba - ca + cb \right)}} \cr & = \left( {{x^0} \times {x^0}} \right) \cr & = \left( {1 \times 1} \right) \cr & = 1 \cr} $$

$$\eqalign{ & {\left( {\sqrt x - \frac{1}{{\sqrt x }}} \right)^2} \cr & = x + \frac{1}{x} - 2 \cr & = \left( {3 + 2\sqrt 2 } \right) + \frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} - 2 \cr & = \left( {3 + 2\sqrt 2 } \right) + \frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{\left( {3 - 2\sqrt 2 } \right)}}{{\left( {3 - 2\sqrt 2 } \right)}} - 2 \cr & = \left( {3 + 2\sqrt 2 } \right) + \left( {3 - 2\sqrt 2 } \right) - 2 \cr & = 4 \cr & \therefore \left( {\sqrt x - \frac{1}{{\sqrt x }}} \right) = 2 \cr} $$

$$\eqalign{ & \frac{{\frac{3}{{2 + \sqrt 3 }} - \frac{2}{{2 - \sqrt 3 }}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{\frac{{3\left( {2 - \sqrt 3 } \right) - 2\left( {2 + \sqrt 3 } \right)}}{{\left( {2 + \sqrt 3 \,} \right)\left( {2 - \sqrt 3 } \right)}}}}{{2 - 5\sqrt 3 }} \cr & = \frac{{6 - 3\sqrt 3 - 4 - 2\sqrt 3 }}{{\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\left( {2 - 5\sqrt 3 } \right)}} \cr & = \frac{{2 - 5\sqrt 3 }}{{2 - 5\sqrt 3 }} \cr & = 1 \cr} $$

$$\eqalign{ & \left( {\sqrt 8 - \sqrt 4 - \sqrt 2 } \right) \cr & = 2\sqrt 2 - 2 - \sqrt 2 \cr & = 2\sqrt 2 - \sqrt 2 - 2 \cr & = \sqrt 2(2 - 1) - 2 \cr & = \sqrt 2 - 2 \cr} $$

$$\eqalign{ & {\text{6}}{{\text{4}}^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {{4^3}} \right)^{ - \frac{2}{3}}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {4^{ - 2}} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^2} \times {\left( {\frac{1}{4}} \right)^{ - 2}} \cr & = {\left( {\frac{1}{4}} \right)^{2 - 2}} \cr & = {\left( {\frac{1}{4}} \right)^0} \cr & = 1 \cr} $$

$$\eqalign{ & \left( {\frac{{{9^2} \times {{18}^4}}}{{{3^{16}}}}} \right) \cr & = \frac{{{9^2} \times {{\left( {9 \times 2} \right)}^4}}}{{{3^{16}}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^2} \times {{\left( {{3^2}} \right)}^4} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^4} \times {3^8} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{\left( {4 + 8} \right)}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{3^{12}} \times {2^4}}}{{{3^{16}}}} \cr & = \frac{{{2^4}}}{{{3^{\left( {16 - 12} \right)}}}} \cr & = \frac{{{2^4}}}{{{3^4}}} \cr & = \frac{{16}}{{81}} \cr} $$

$$\eqalign{ & \frac{{{4^3} \times {5^4}}}{{{4^5}}} \cr & = \frac{{{5^4}}}{{{4^{\left( {5 - 3} \right)}}}} \cr & = \frac{{{5^4}}}{{{4^2}}} \cr & = \frac{{625}}{{16}} \cr & = 39.0625 \cr} $$

$$\eqalign{ & \frac{{{9^3} \times {6^2}}}{{{3^3}}} \cr & = \frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {3 \times 2} \right)}^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {3 \times 2} \right)}} \times {3^2} \times {2^2}}}{{{3^3}}} \cr & = \frac{{{3^{\left( {6 + 2} \right)}} \times {2^2}}}{{{3^3}}} \cr & = {3^{\left( {8 - 3} \right)}} \times {2^2} \cr & = {3^5} \times {2^2} \cr & = 243 \times 4 \cr & = 972 \cr} $$

$$\left( {\frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} + \frac{{2 - \sqrt 3 }}{{2 + \sqrt 3 }} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}} \right)$$ $$ = \left\{ {\frac{{{{\left( {2 + \sqrt 3 } \right)}^2} + {{\left( {2 - \sqrt 3 } \right)}^2}}}{{\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}} + \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}} \right\}$$ $$ = \left\{ {\frac{{4 + 3 + 4\sqrt 3 + 4 + 3 - 4\sqrt 3 }}{{4 - 3}} + \frac{{{{\left( {\sqrt 3 - 1} \right)}^2}}}{{3 - 1}}} \right\}$$ $$\eqalign{ & = \left\{ {14 + \frac{{3 + 1 - 2\sqrt 3 }}{2}} \right\} \cr & = 14 + \frac{{2\left( {2 - \sqrt 3 } \right)}}{2} \cr & = 14 + 2 - \sqrt 3 \cr & = 16 - \sqrt 3 \cr} $$

$$\eqalign{ & \sqrt {\frac{{\left( {\sqrt {12} - \sqrt 8 } \right)\left( {\sqrt 3 + \sqrt 2 } \right)}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{\sqrt {36} + \sqrt {24} - \sqrt {24} - \sqrt {16} }}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{{6 - 4}}{{5 + \sqrt {24} }}} \cr & = \sqrt {\frac{2}{{5 + \sqrt {24} }} \times \frac{{5 - \sqrt {24} }}{{5 - \sqrt {24} }}} \cr & = \sqrt {\frac{{2\left( {5 - \sqrt {24} } \right)}}{{25 - 24 }}} \cr & = \sqrt {2\left( {5 - 2\sqrt 6 } \right)} \cr & = \sqrt {2\left\{ {{{\left( {\sqrt 3 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 3 \times \sqrt 2 } \right\}} \cr & = \sqrt {2{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} \cr & = \sqrt 2 \left( {\sqrt 3 - \sqrt 2 } \right) \cr & = \sqrt 6 - 2 \cr} $$

$$\eqalign{ & \frac{{{{\left( {19} \right)}^{12}} \times {{\left( {19} \right)}^8}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{19}^{\left( {12 + 8} \right)}}}}{{{{\left( {19} \right)}^4}}} \cr & = \frac{{{{\left( {19} \right)}^{20}}}}{{{{\left( {19} \right)}^4}}} \cr & = {\left( {19} \right)^{\left( {20 - 4} \right)}} \cr & = {\left( {19} \right)^{16}} \cr} $$ Hence the missing number = 16

$$\eqalign{ & {\left( {64} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( {{8^2}} \right)^4} \div {\left( 8 \right)^5} \cr & = {\left( 8 \right)^{\left( {2 \times 4} \right)}} \div {8^5} \cr & = \frac{{{8^8}}}{{{8^5}}} \cr & = {8^{\left( {8 - 5} \right)}} \cr & = {8^3} \cr} $$

$$\frac{1}{{\sqrt {\left( {12 - \sqrt {140} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {60} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {84} } \right)} }}$$ $$ = \frac{1}{{\sqrt {\left( {12 - \sqrt {4 \times 35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - \sqrt {4 \times 15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + \sqrt {4 \times 21} } \right)} }}$$ $$ = \frac{1}{{\sqrt {\left( {12 - 2\sqrt {35} } \right)} }} - \frac{1}{{\sqrt {\left( {8 - 2\sqrt {15} } \right)} }} - \frac{2}{{\sqrt {\left( {10 + 2\sqrt {21} } \right)} }}$$ $$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 7 .\sqrt 5 } }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} - 2.\sqrt 5 .\sqrt 3 } }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + {{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 7 .\sqrt 3 } }}$$ $$ = \frac{1}{{\sqrt {{{\left( {\sqrt 7 - \sqrt 5 } \right)}^2}} }} - \frac{1}{{\sqrt {{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}} }} - \frac{2}{{\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} }}$$ $$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }}$$ Rationalizing in above equation, $$ = \frac{1}{{\sqrt 7 - \sqrt 5 }} \times \frac{{\sqrt 7 + \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} - \frac{1}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} - \frac{2}{{\sqrt 7 + \sqrt 3 }} \times \frac{{\sqrt 7 - \sqrt 3 }}{{\sqrt 7 - \sqrt 3 }}$$ $$ = \frac{{\sqrt 7 + \sqrt 5 }}{2} - \frac{{\sqrt 5 + \sqrt 3 }}{2} - \frac{{\sqrt 7 - \sqrt 3 }}{2}$$ $$ = \frac{{\sqrt 7 + \sqrt 5 - \sqrt 5 - \sqrt 3 - \sqrt 7 + \sqrt 3 }}{2}$$ $$ = 0$$

$$\eqalign{ & \frac{{{{\left( {1000} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {{{10}^3}} \right)}^{12}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{\left( {3 \times 12} \right)}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = \frac{{{{\left( {10} \right)}^{36}}}}{{{{\left( {10} \right)}^{30}}}} \cr & = {\left( {10} \right)^{\left( {36 - 30} \right)}} \cr & = {10^6} \cr & = {\left( {{{10}^3}} \right)^2} \cr & = {\left( {1000} \right)^2} \cr} $$

$$\frac{{{{1.5}^3} + {{4.7}^3} + {{3.8}^3} - 3 \times 1.5 \times 4.7 \times 3.8}}{{{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5}}$$ $$ = \frac{{\left( {1.5 + 4.7 + 3.8} \right)\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}{{\left\{ {{{1.5}^2} + {{4.7}^2} + {{3.8}^2} - 1.5 \times 4.7 - 4.7 \times 3.8 - 3.8 \times 1.5} \right\}}}$$ $$\left[ {\therefore {a^3} + {b^3} + {c^3} - 3abc = \left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right)} \right]$$ $$\eqalign{ & = 1.5 + 4.7 + 3.8 \cr & = 10.0 \cr & = 10 \cr} $$

$$\eqalign{ & \frac{{0.41 \times 0.41 \times 0.41 + 0.69 \times 0.69 \times 0.69}}{{0.41 \times 0.41 - 0.41 \times 0.69 + 0.69 \times 0.69}} \cr & = \frac{{{{\left( {0.41} \right)}^3} + {{\left( {0.69} \right)}^3}}}{{{{\left( {0.41} \right)}^2} - 0.41 \times 0.69 + {{\left( {0.69} \right)}^2}}} \cr & .....\left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = 0.41 + 0.69 \cr & = 1.1 \cr} $$

$$\eqalign{ & {{\text{3}}^{10}} \times {\text{2}}{{\text{7}}^2}{\text{ = }}{{\text{9}}^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10}} \times {\left( {{3^3}} \right)^2}{\text{ = }}{\left( {{3^2}} \right)^2} \times {{\text{3}}^n} \cr & \Rightarrow {{\text{3}}^{10 + 6}}{\text{ = }}{{\text{3}}^{4 + n}} \cr & \Rightarrow 10 + 6 = 4 + n \cr & \Rightarrow n = 12 \cr} $$

3 8 × 3 4 = 3 (8 + 4) = 3 12 = (3 6 ) 2 = (729) 2

$$\eqalign{ & \frac{{343 \times 49}}{{216 \times 16 \times 81}} \cr & = \frac{{{7^3} \times {7^2}}}{{{6^3} \times {2^4} \times {3^4}}} \cr & = \frac{{{7^{\left( {3 + 2} \right)}}}}{{{6^3} \times {{\left( {2 \times 3} \right)}^4}}} \cr & = \frac{{{7^5}}}{{{6^3} \times {6^4}}} \cr & = \frac{{{7^5}}}{{{6^{\left( {3 + 4} \right)}}}} \cr & = \frac{{{7^5}}}{{{6^7}}} \cr} $$

$$\eqalign{ & \frac{{16 \times 32}}{{9 \times 27 \times 81}} \cr & = \frac{{{2^4} \times {2^5}}}{{{3^2} \times {3^3} \times {3^4}}} \cr & = \frac{{{2^{\left( {4 + 5} \right)}}}}{{{3^{\left( {2 + 3 + 4} \right)}}}} \cr & = \frac{{{2^9}}}{{{3^9}}} \cr & = {\left( {\frac{2}{3}} \right)^9} \cr} $$

$$\eqalign{ & \sqrt 2 = 1.414 \cr & \Rightarrow \sqrt 8 {\text{ + 2}}\sqrt {32} - 3\sqrt {128} {\text{ + 4}}\sqrt {50} \cr & \Rightarrow 2\sqrt 2 + 2 \times 4\sqrt 2 - 3 \times 8\sqrt 2 + 4 \times 5\sqrt 2 \cr & \Rightarrow 2\sqrt 2 + 8\sqrt 2 - 24\sqrt 2 + 20\sqrt 2 \cr & \Rightarrow 6\sqrt 2 \cr & \Rightarrow 6 \times 1.414 \cr & \Rightarrow 8.484{\text{ }} \cr} $$

$$\eqalign{ & {\text{Let}}\,{9^3} \times {\left( {81} \right)^2} \div {\left( {27} \right)^3} = {\left( 3 \right)^x}{\text{then}} \cr & \Rightarrow {\left( 3 \right)^x}{\text{ = }}\frac{{{{\left( {{3^2}} \right)}^3} \times {{\left( {{3^4}} \right)}^2}}}{{{{\left( {{3^3}} \right)}^3}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {2 \times 3} \right)}} \times {3^{\left( {4 \times 2} \right)}}}}{{{3^{\left( {3 \times 3} \right)}}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^6} \times {3^8}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{\left( {6 + 8} \right)}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = \frac{{{3^{14}}}}{{{3^9}}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^{\left( {14 - 9} \right)}} \cr & \Rightarrow {\left( 3 \right)^x} = {3^5} \cr & \Rightarrow {\left( 3 \right)^x} = 5 \cr} $$

$$\eqalign{ & {\text{Let ,}} \cr & {\left( 6 \right)^4} \div {\left( {36} \right)^3} \times 216 = {6^{\left( {x - 5} \right)}} \cr & {\text{Then,}} \cr & {6^{\left( {x - 5} \right)}} = {\left( 6 \right)^4} \div {\left( {{6^2}} \right)^3} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^{\left( {2 \times 3} \right)}} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^4} \div {6^6} \times {6^3} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = {6^{\left( {4 - 6 + 3} \right)}} \cr & \Rightarrow {6^{\left( {x - 5} \right)}} = 6 \cr & \Rightarrow x - 5 = 1 \cr & \Rightarrow x = 6 \cr} $$

$$\eqalign{ & {\left( {256} \right)^{\frac{5}{4}}} \cr & = {\left( {{4^4}} \right)^{\frac{5}{4}}} \cr & = {4^{\left( {4 \times \frac{5}{4}} \right)}} \cr & = {4^5} \cr & = 1024 \cr} $$

$$\eqalign{ & x = \sqrt {2 + \sqrt {2 + \sqrt {2 + ......} } } \cr & \Rightarrow {x^2} = 2 + \sqrt {2 + \sqrt {2 + .......} } \cr & \Rightarrow {x^2} = 2 + x \cr & \Rightarrow {x^2} - x - 2 = 0 \cr & \Rightarrow x\left( {x - 2} \right) + 1\left( {x - 2} \right) = 0 \cr & \Rightarrow \left( {x + 1} \right)\left( {x - 2} \right) = 0 \cr & \Rightarrow x = 2 \cr} $$

$$\eqalign{ & {\text{2 + }}\sqrt {0.09} - \root 3 \of {0.008} - 75\% \,{\text{of }}2.80 \cr & = 2 + 0.3 - 0.2 - \left( {\frac{3}{4} \times 2.8} \right) \cr & = 2 + 0.3 - 0.2 - 2.10 \cr & = 2.3 - 2.3 \cr & = 0 \cr} $$

$$\eqalign{ & {\left( {3 + 2\sqrt 2 } \right)^{ - 3}} + {\left( {3 - 2\sqrt 2 } \right)^{ - 3}} \cr & = {\left( {\frac{1}{{3 + 2\sqrt 2 }}} \right)^3} + {\left( {\frac{1}{{3 - 2\sqrt 2 }}} \right)^3} \cr} $$ $$ = {\left( {\frac{1}{{\left( {3 - 2\sqrt 2 } \right)}} \times \frac{{3 + 2\sqrt 2 }}{{3 + 2\sqrt 2 }}} \right)^3} + $$ $${\left( {\frac{1}{{\left( {3 + 2\sqrt 2 } \right)}} \times \frac{{3 - 2\sqrt 2 }}{{3 - 2\sqrt 2 }}} \right)^3}$$ $$\eqalign{ & = {\left( {\frac{{3 - 2\sqrt 2 }}{{9 - 8}}} \right)^3} + {\left( {\frac{{3 + 2\sqrt 2 }}{{9 - 8}}} \right)^3} \cr & = {\left( {3 - 2\sqrt 2 } \right)^3} + {\left( {3 + 2\sqrt 2 } \right)^3} \cr & a = 3 - 2\sqrt 2 \cr & b = 3 + 2\sqrt 2 \cr & \left[ {\because {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)} \right] \cr & = \left( {3 - 2\sqrt 2 + 3 + 2\sqrt 2 } \right)\left( {17 + 17 - 1} \right) \cr & = \left( 6 \right)\left( {33} \right) \cr & = 198 \cr} $$