Surds And Indices — Part 2 MCQs with Explanations

$$\eqalign{ & {\left( {\sqrt 8 } \right)^{\frac{1}{3}}} \cr & = {\left( {{8^{\frac{1}{2}}}} \right)^{\frac{1}{3}}} \cr & = {8^{\left( {\frac{1}{2} \times \frac{1}{3}} \right)}} \cr & = {8^{\frac{1}{6}}} \cr & = {\left( {{2^3}} \right)^{\frac{1}{6}}} \cr & = {2^{\left( {3 \times \frac{1}{6}} \right)}} \cr & = {2^{\frac{1}{2}}} \cr & = \sqrt 2 \cr} $$

$$\eqalign{ & {\left( {\frac{{32}}{{243}}} \right)^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left\{ {{{\left( {\frac{2}{3}} \right)}^5}} \right\}^{ - \frac{4}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{5 \times \frac{{\left( { - 4} \right)}}{5}}} \cr & {\text{ = }}{\left( {\frac{2}{3}} \right)^{\left( { - 4} \right)}} \cr & {\text{ = }}{\left( {\frac{3}{2}} \right)^4} \cr & {\text{ = }}\frac{{{3^4}}}{{{2^4}}} \cr & {\text{ = }}\frac{{81}}{{16}}{\text{ }} \cr} $$

$$\eqalign{ & {\text{ 2}}{{\text{7}}^{ - \frac{2}{3}}} \cr & = {\left( {{3^3}} \right)^{ - \frac{2}{3}}} \cr & = {3^{\left[ {3 \times \left( { - \frac{2}{3}} \right)} \right]}} \cr & = {3^{ - 2}} \cr & = \frac{1}{{{3^2}}} \cr & = \frac{1}{9} \cr & {\text{Clearly}},\,\,0 < \frac{1}{9} < 1 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt {3.25} + \sqrt {2.25} }} \times \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{\sqrt {3.25} - \sqrt {2.25} }} \cr & = \frac{{\sqrt {3.25} - \sqrt {2.25} }}{{3.25 - 2.25}} \cr & = \sqrt {3.25} - \sqrt {2.25} \,......(i) \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt {4.25} + \sqrt {3.25} }} \cr & = \sqrt {4.25} - \sqrt {3.25} \,.......(ii) \cr & \frac{1}{{\sqrt {5.25} + \sqrt {4.25} }} \cr & = \sqrt {5.25} - \sqrt {4.25} \,.......(iii) \cr & \frac{1}{{\sqrt {6.25} + \sqrt {5.25} }} \cr & = \sqrt {6.25} - \sqrt {5.25} \,.......(iv) \cr & {\text{Now}}\,{\text{add}}\,{\text{all}}\,{\text{them}} \cr & \sqrt {3.25} - \sqrt {2.25} + \sqrt {4.25} - \sqrt {3.25} + \sqrt {5.25} - \sqrt {4.25} + \sqrt {6.25} - \sqrt {5.25} \cr & = \sqrt {6.25} - \sqrt {2.25} \cr & = 2.5 - 1.5 \cr & = 1 \cr} $$

$$\eqalign{ & \frac{{{3^0} + {3^{ - 1}}}}{{{3^{ - 1}} - {3^0}}} \cr & = \frac{{1 + \frac{1}{3}}}{{\frac{1}{3} - 1}} \cr & = \frac{{\frac{4}{3}}}{{ - \frac{2}{3}}} \cr & = - 2 \cr} $$

$$\frac{2}{{\sqrt 7 + \sqrt 5 }} + \frac{7}{{\sqrt {12} - \sqrt 5 }} - \frac{5}{{\sqrt {12} - \sqrt 7 }}$$ $$ = \frac{2}{{\sqrt 7 + \sqrt 5 }} \times $$ $$\frac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 - \sqrt 5 }} + $$ $$\frac{7}{{\sqrt {12} - \sqrt 5 }} \times $$ $$\frac{{\sqrt {12} + \sqrt 5 }}{{\sqrt {12} + \sqrt 5 }} - $$ $$\left( {\frac{5}{{\sqrt {12} - \sqrt 7 }} \times \frac{{\sqrt {12} + \sqrt 7 }}{{\sqrt {12} + \sqrt 7 }}} \right)$$ $$ = \frac{{2\left( {\sqrt 7 - \sqrt 5 } \right)}}{2} + $$ $$\frac{{7\left( {\sqrt {12} + \sqrt 5 } \right)}}{7} - $$ $$\frac{{5\left( {\sqrt {12} + \sqrt 7 } \right)}}{5}$$ $$\eqalign{ &= \sqrt 7 - \sqrt 5 + \sqrt {12} + \sqrt 5 - \sqrt {12} - \sqrt 7 \cr &= 0 \cr} $$

$$\eqalign{ & \root 3 \of {{2^4}\sqrt {{2^{ - 5}}\sqrt {{2^6}} } } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( {{2^6}} \right)}^{\frac{1}{2}}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( 2 \right)}^{\left( {6 \times \frac{1}{2}} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{.2}^3}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{\left( { - 5 + 3} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 2}}} } {\text{ }} \cr & = \root 3 \of {{2^4}.{{\left( {{2^2}} \right)}^{\frac{1}{2}}}} \cr & {\text{ = }}\root 3 \of {{2^4}{{.2}^{\left( { - 2 \times \frac{1}{2}} \right)}}} \cr & = \root 3 \of {{2^4}{{.2}^{\left( { - 1} \right)}}} \cr & = \root 3 \of {{2^{\left( {4 - 1} \right)}}} \cr & {\text{ = }}\root 3 \of {{2^3}} \cr & {\text{ = }}{\left( {{2^3}} \right)^{\frac{1}{3}}} \cr & {\text{ = }}{{\text{2}}^{\left( {3 \times \frac{1}{3}} \right)}} \cr & {\text{ = 2 }} \cr} $$

$$\eqalign{ & \sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt 2 } } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{1}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{1}{2}} \right)}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2\sqrt {{2^{\frac{3}{2}}}} } } } \cr & = \sqrt {2\sqrt {2\sqrt {2.{{\left( {{2^{\frac{3}{2}}}} \right)}^{\frac{1}{2}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\left( {\frac{3}{2} \times \frac{1}{2}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{{2.2}^{\frac{3}{4}}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\left( {1 + \frac{3}{4}} \right)}}} } } \cr & = \sqrt {2\sqrt {2\sqrt {{2^{\frac{7}{4}}}} } } \cr & = \sqrt {2\sqrt {2{{\left( {{2^{\frac{7}{4}}}} \right)}^{\frac{1}{2}}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\left( {\frac{7}{4} \times \frac{1}{2}} \right)}}} } \cr & = \sqrt {2\sqrt {{{2.2}^{\frac{7}{8}}}} } \cr & = \sqrt {2\sqrt {{2^{\left( {1 + \frac{7}{8}} \right)}}} } \cr & = \sqrt {2\sqrt {{2^{\frac{{15}}{8}}}} } \cr & = \sqrt {2{{\left( {{2^{\frac{{15}}{8}}}} \right)}^{\frac{1}{2}}}} \cr & = \sqrt {{{2.2}^{\frac{{15}}{{16}}}}} \cr & = \sqrt {{2^{\left( {1 + \frac{{15}}{{16}}} \right)}}} \cr & = \sqrt {{2^{\frac{{31}}{{16}}}}} \cr & = {\left( {{2^{\frac{{31}}{{16}}}}} \right)^{\frac{1}{2}}} \cr & = {2^{\left( {\frac{{31}}{{16}} \times \frac{1}{2}} \right)}} \cr & = {2^{\frac{{31}}{{32}}}} \cr} $$

$$\eqalign{ & {\left( {0.03125} \right)^{ - \frac{2}{5}}} \cr & = {\left[ {{{\left( {0.5} \right)}^5}} \right]^{ - \frac{2}{5}}} \cr & = {0.5^{\left[ {5 \times \left( { - \frac{2}{5}} \right)} \right]}} \cr & = {\left( {0.5} \right)^{ - 2}} \cr & = \frac{1}{{{{\left( {0.5} \right)}^2}}} \cr & = \frac{1}{{0.25}} \cr & = 4 \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 3 + \sqrt 4 }} \cr & = \frac{1}{{\sqrt 4 + \sqrt 3 }} \times \frac{{\sqrt 4 - \sqrt 3 }}{{\sqrt 4 - \sqrt 3 }} \cr & = \frac{{\sqrt 4 - \sqrt 3 }}{1} \cr & = \sqrt 4 - \sqrt 3 \cr & {\text{Similarly}} \cr & \frac{1}{{\sqrt 4 + \sqrt 5 }} = \sqrt 5 - \sqrt 4 \cr & \Rightarrow \frac{1}{{\sqrt 5 + \sqrt 6 }} = \sqrt 6 - \sqrt 5 \cr & \Rightarrow \frac{1}{{\sqrt 6 + \sqrt 7 }} = \sqrt 7 - \sqrt 6 \cr & \Rightarrow \frac{1}{{\sqrt 7 + \sqrt 8 }} = \sqrt 8 - \sqrt 7 \cr & \Rightarrow \frac{1}{{\sqrt 8 + \sqrt 9 }} = \sqrt 9 - \sqrt 8 \cr & {\text{Now put values}} \cr & \Rightarrow \sqrt 4 - \sqrt 3 + \sqrt 5 - \sqrt 4 + \sqrt 6 - \sqrt 5 + \sqrt 7 - \sqrt 6 + \sqrt 8 - \sqrt 7 + \sqrt 9 - \sqrt 8 \cr & \Rightarrow \sqrt 9 - \sqrt 3 \cr & \Rightarrow 3 - \sqrt 3 \cr} $$

$$\eqalign{ & {\text{let}} \cr & a = 0.73 \cr & b = 0.27 \cr & = \frac{{{a^3} + {b^3}}}{{{a^2} + {b^2} - ab}} \cr & = \frac{{\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)}}{{\left( {{a^2} + {b^2} - ab} \right)}} \cr & = \left( {a + b} \right) \cr & = \left( {0.73 + 0.27} \right) \cr & = 1 \cr} $$

$$\eqalign{ & x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & \Rightarrow x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \times \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & \Rightarrow x = \frac{{{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Similarly}} \cr & y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }} \cr & \Rightarrow y = \frac{{{{\left( {\sqrt 5 - \sqrt 3 } \right)}^2}}}{2} \cr & {\text{Now, }}x + y \cr & = \frac{{5 + 3 + 2\sqrt {15} + 5 + 3 - 2\sqrt {15} }}{2} \cr & = \frac{{16}}{2} \cr & = 8 \cr} $$

$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left\{ {{{\left( {{x^{ - \frac{3}{5}}}} \right)}^{\frac{1}{5}}}} \right\}}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{^{\left\{ {\left( { - \frac{3}{5}} \right) \times \frac{1}{5}} \right\}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^5} \cr & = {\left[ {{x^{\left\{ {\left( { - \frac{3}{{25}}} \right) \times \left( { - \frac{5}{3}} \right)} \right\}}}} \right]^5} \cr & = {\left( {{x^{\frac{1}{5}}}} \right)^5} \cr & = {x^{\left( {\frac{1}{5} \times 5} \right)}} \cr & = x \cr} $$

$$\eqalign{ & {\text{Let }}\frac{{{{\left( x \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( x \right)}^{\frac{1}{3}}}}} \cr & {\text{Then,}} \cr & {x^{\frac{2}{3}}}.{x^{\frac{1}{3}}} = 42 \times 5 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 210 \cr & \Rightarrow {x^{\left( {\frac{2}{3} + \frac{1}{3}} \right)}} = 210 \cr & \Rightarrow x = 210 \cr} $$

$$\eqalign{ & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}} \cr & = {5^{0.25}} \times {\left( {{5^3}} \right)^{0.25}} \cr & = {5^{0.25}} \times {5^{\left( {3 \times 0.25} \right)}} \cr & = {5^{0.25}} \times {5^{0.75}} \cr & = {5^{\left( {0.25 + 0.75} \right)}} \cr & = {5^1} \cr & = 5 \cr} $$

$$\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\left[ {\left( {3 + \sqrt 5 } \right) + 2\sqrt 2 } \right]\left[ {\left( {3 + \sqrt 5 } \right) - 2\sqrt 2 } \right]}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{{{\left( {3 + \sqrt 5 } \right)}^2} - {{\left( {2\sqrt 2 } \right)}^2}}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{9 + 5 + 6\sqrt 5 - 8}} \cr & = \frac{{12\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{6\sqrt 5 + 6}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)}}{{\sqrt 5 + 1}} \cr & = \frac{{2\left( {3 + \sqrt 5 - 2\sqrt 2 } \right)\left( {\sqrt 5 - 1} \right)}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left( {3\sqrt 5 + 5 - 2\sqrt {10} - 3 - \sqrt 5 + 2\sqrt 2 } \right)}}{{5 - 1}} \cr & = \frac{{2\left( {2\sqrt 5 + 2\sqrt 2 - 2\sqrt {10} + 2} \right)}}{4} \cr & = \frac{{2 \times 2\left( {\sqrt 5 + \sqrt 2 - \sqrt {10} + 1} \right)}}{4} \cr & = \sqrt 5 + \sqrt 2 - \sqrt {10} + 1 \cr & {\text{or,}}\,1 + \sqrt 5 + \sqrt 2 - \sqrt {10} \cr} $$

$$\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{2 \times \frac{9}{4}}}\sqrt {{2^{1 + 2}}} }}{{2\sqrt {\frac{1}{4}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {\frac{{{2^{\frac{9}{2}}}{{.2}^{\frac{3}{2}}}}}{{2 \times \frac{1}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - {{\left( {{2^{\frac{{12}}{2}}}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & = \left[ {8 - \left( {{2^{6 \times \frac{1}{2}}}} \right)} \right] \cr & = \left[ {8 - \left( {{2^3}} \right)} \right] \cr & = \left[ {8 - 8} \right] \cr & = 0 \cr} $$

$$\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} - \frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} + \frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}$$ $$ = \left( {\frac{{3\sqrt 2 }}{{\sqrt 6 + \sqrt 3 }} \times \frac{{\sqrt 6 - \sqrt 3 }}{{\sqrt 6 - \sqrt 3 }}{\text{ }}} \right) - $$ $$\left( {\frac{{2\sqrt 6 }}{{\sqrt 3 + 1}} \times \frac{{\sqrt 3 - 1}}{{\sqrt 3 - 1}}{\text{ }}} \right) + $$ $$\left( {\frac{{2\sqrt 3 }}{{\sqrt 6 + 2}}{\text{ }} \times \frac{{\sqrt 6 - 2}}{{\sqrt 6 - 2}}{\text{ }}} \right)$$ $$ = \frac{{3\sqrt 2 \left( {\sqrt 6 - \sqrt 3 } \right)}}{3} - $$ $$\frac{{2\sqrt 6 \left( {\sqrt 3 - 1} \right)}}{2} + $$ $$\frac{{2\sqrt 3 \left( {\sqrt 6 - 2} \right)}}{2}$$ $$\eqalign{ & = \sqrt {12} - \sqrt 6 - \sqrt {18} + \sqrt 6 + \sqrt {18} - 2\sqrt 3 \cr & = \sqrt {12} - 2\sqrt 3 \cr & = 2\sqrt 3 - 2\sqrt 3 \cr & = 0 \cr} $$

$$\eqalign{ & \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{{\left( {{6^3}} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {{4^4}} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {{2^5}} \right)}^{ - \frac{1}{5}}}}} \cr & = \frac{1}{{{6^{3 \times \frac{{\left( { - 2} \right)}}{3}}}}}{\text{ + }}\frac{1}{{{4^{4 \times \frac{{\left( { - 3} \right)}}{4}}}}} + \frac{1}{{{2^{5 \times \frac{{\left( { - 1} \right)}}{5}}}}} \cr & = \frac{1}{{{6^{ - 2}}}}{\text{ + }}\frac{1}{{{4^{ - 3}}}}{\text{ + }}\frac{1}{{{2^{ - 1}}}} \cr & = \left( {{6^2} + {4^3} + {2^1}} \right) \cr & = \left( {36 + 64 + 2} \right) \cr & = 102 \cr} $$

$$\eqalign{ & {\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16 \times 3} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{ - \frac{2}{7}}} \times {\left( 3 \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} \cr & = {\left( {16} \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{2}{7} - \frac{5}{7}} \right)}} \cr & = {\left( {16} \right)^{\left( { - \frac{7}{7}} \right)}} \times {\left( 3 \right)^{\left( { - \frac{7}{7}} \right)}} \cr & = {\left( {16} \right)^{ - 1}} \times {\left( 3 \right)^{ - 1}} \cr & = \frac{1}{{16}} \times \frac{1}{3} \cr & = \frac{1}{{48}} \cr} $$

$$\eqalign{ & {\text{1}}{{\text{0}}^{ - 8x}} \cr & = {\left[ {{{10}^x}} \right]^{ - 8}} \cr & = {\left[ {\frac{1}{2}} \right]^{ - 8}} \cr & = {2^8} \cr & = 256 \cr} $$

$$\left( {\frac{1}{{1.4}} + \frac{1}{{4.7}} + \frac{1}{{7.10}} + \frac{1}{{10.13}} + \frac{1}{{13.16}}} \right)$$ Formula : $$\frac{1}{{{\text{Difference denominator value}}}} \times $$ $$\left[ {\frac{1}{{{\text{First value}}}} - \frac{1}{{{\text{Last value}}}}} \right]$$ $$ = \frac{1}{3} \times $$ $$\left[ {1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{{10}} + \frac{1}{{10}} - \frac{1}{{13}} + \frac{1}{{13}} - \frac{1}{{16}}} \right]$$ $$\eqalign{ & = \frac{1}{3} \times \left[ {1 - \frac{1}{{16}}} \right] \cr & = \frac{1}{3} \times \frac{{15}}{{16}} \cr & = \frac{5}{{16}} \cr} $$

$$\eqalign{ & \frac{1}{{\sqrt 5 + \sqrt 3 }} \times \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }} \cr & = \frac{{\sqrt 5 - \sqrt 3 }}{{5 - 3}} \cr & = \frac{{2.236 - 1.732}}{2} \cr & = \frac{{0.504}}{2} \cr & = 0.252 \cr} $$

$$\eqalign{ & {\text{ }}{\left( {\frac{3}{5}} \right)^3}{\left( {\frac{3}{5}} \right)^{ - 6}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{\left( {3 - 6} \right)}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow {\text{ }}{\left( {\frac{3}{5}} \right)^{ - 3}} = {\left( {\frac{3}{5}} \right)^{2x - 1}} \cr & \Rightarrow 2x - 1 = - 3 \cr & \Rightarrow 2x = - 2 \cr & \Rightarrow x = - 1 \cr} $$

$$\eqalign{ & {\text{5}}\sqrt 5 \times {5^3} \div {5^{ - \frac{3}{2}}}{\text{ = }}{{\text{5}}^{a + 2}} \cr & \Rightarrow \frac{{5 \times {5^{\frac{1}{2}}} \times {5^3}}}{{{5^{ - \frac{3}{2}}}}} = {5^{a + 2}} \cr & \Rightarrow {5^{\left( {1 + \frac{1}{2} + 3 + \frac{3}{2}} \right)}} = {5^{a + 2}} \cr & \Rightarrow {5^6} = {5^{a + 2}} \cr & \Rightarrow a + 2 = 6 \cr & \Rightarrow a = 4 \cr} $$

$$\eqalign{ & {{\text{2}}^x} = \root 3 \of {32} \cr & \Rightarrow {{\text{2}}^x} = {\left( {32} \right)^{\frac{1}{3}}} = {\left( {{2^5}} \right)^{\frac{1}{3}}} \cr & \Rightarrow x = \frac{5}{3} \cr} $$

$$\sqrt {12 + \sqrt {12 + \sqrt {12 + ..... } } } $$ (3, 4) are the factor of 4 If there is '+' in $$\sqrt {\,\,\,} $$ Answer is Highest value If there is '-' in $$\sqrt {\,\,\,} $$ Answer is lowest value. So the answer is 4

$$\eqalign{ & {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = 2}}{{\text{5}}^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{\left( {{5^2}} \right)^{(3x - 4)}} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^2}^{(3x - 4)} \cr & \Rightarrow {{\text{5}}^{\left( {x + 3} \right)}}{\text{ = }}{{\text{5}}^{(6x - 8)}} \cr & \Rightarrow x + 3 = 6x - 8 \cr & \Rightarrow 5x = 11 \cr & \Rightarrow x = \frac{{11}}{5} \cr} $$

$$\eqalign{ & \frac{{{2^{n + 4}} - 2\left( {{2^n}} \right)}}{{2\left( {{2^{n + 3}}} \right)}}{\text{ }} \cr & = \frac{{{2^{n + 4}} - {2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = \frac{{{2^{n + 4}}}}{{{2^{n + 4}}}} - \frac{{{2^{n + 1}}}}{{{2^{n + 4}}}}{\text{ }} \cr & = 1 - {2^{(n + 1) - \left( {n + 4} \right)}} \cr & {\text{ = 1}} - {2^{ - 3}} \cr & {\text{ = 1}} - \frac{1}{8}{\text{ }} \cr & {\text{ = }}\frac{7}{8} \cr} $$

$$\eqalign{ & a + b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} + \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} \cr & = \frac{{{{\left( {\sqrt 5 + 1} \right)}^2} + {{\left( {\sqrt 5 - 1} \right)}^2}}}{{\left( {\sqrt 5 - 1} \right)\left( {\sqrt 5 + 1} \right)}} \cr & = \frac{{2\left[ {{{\left( {\sqrt 5 } \right)}^2} + 1} \right]}}{{5 - 1}} \cr & = \frac{{2\left( {5 + 1} \right)}}{4} \cr & = 3 \cr & a.b = \frac{{\sqrt 5 + 1}}{{\sqrt 5 - 1}} \times \frac{{\sqrt 5 - 1}}{{\sqrt 5 + 1}} = 1 \cr & {\text{Put value in expression}} \cr & \frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} \cr & = \frac{{{{\left( {a + b} \right)}^2} - ab}}{{{{\left( {a + b} \right)}^2} - 3ab}} \cr & = \frac{{{3^2} - 1}}{{{3^2} - 3}} \cr & = \frac{{9 - 1}}{{9 - 3}} \cr & = \frac{4}{3} \cr} $$

$$\eqalign{ & {\text{Let }}\,{6^{1.2}} \times {36^x} \times {30^{2.4}} \times {25^{1.3}} = {30^5} \cr & {\text{Then,}}\,{6^{1.2}} \times {({6^2})^x} \times {(6 \times 5)^{2.4}} \times {({5^2})^{1.3}} = {30^5} \cr & \Leftrightarrow {6^{1.2}} \times {6^{2x}} \times {6^{2.4}} \times {5^{2.4}} \times {5^{2.6}} = {(6 \times 5)^5} \cr & \Leftrightarrow {6^{\left( {1.2 + 2x + 2.4} \right)}} \times {5^{\left( {2.4 + 2.6} \right)}} = {6^5} \times {5^5} \cr & \Leftrightarrow {6^{\left( {3.6 + 2x} \right)}} \times {5^5} = {6^5} \times {5^5} \cr & \Leftrightarrow 3.6 + 2x = 5 \cr & \Leftrightarrow 2x = 1.4 \cr & \Leftrightarrow x = 0.7 \cr} $$

$$\left[ {\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} - \frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }}} \right]$$ $$ = \frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 + \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} - $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 + \sqrt 2 }} \times $$ $$\frac{{\sqrt 3 - \sqrt 2 }}{{\sqrt 3 - \sqrt 2 }}$$ $$\eqalign{ & = \frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{{3 - 2}} - \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{{3 - 2}} \cr & = \left( {3 + 2 + 2\sqrt 6 } \right) - \left( {3 + 2 - 2\sqrt 6 } \right) \cr & = 4\sqrt 6 {\text{ }} \cr} $$

$$\eqalign{ & \sqrt 3 = 1.732 \cr & \Rightarrow \frac{{2 + \sqrt 3 }}{{2 - \sqrt 3 }} \times \frac{{2 + \sqrt 3 }}{{2 + \sqrt 3 }} \cr & \Rightarrow \frac{{{{\left( {2 + \sqrt 3 } \right)}^2}}}{{4 - 3}} \cr & \Rightarrow 4 + 3 + 4\sqrt 3 \cr & \Rightarrow 7 + 4 \times 1.732 \cr & \Rightarrow 7 + 6.928 \cr & \Rightarrow 13.928 \cr} $$

$$\eqalign{ & 16\sqrt {\frac{3}{4}} - 9\sqrt {\frac{4}{3}} \cr & \Rightarrow 16\sqrt {\frac{{3 \times 4}}{{4 \times 4}}} - 9\sqrt {\frac{{3 \times 4}}{{3 \times 3}}} \cr & \Rightarrow 16 \times \frac{{\sqrt {12} }}{4} - \frac{{9\sqrt {12} }}{3} \cr & \Rightarrow 4\sqrt {12} - 3\sqrt {12} \cr & \Rightarrow \sqrt {12} \cr & \Rightarrow 3.46 \cr} $$

$$\eqalign{ & {\text{Let }}{2^{3.6}} \times {4^{3.6}} \times {4^{3.6}} \times {(32)^{2.3}} = {\left( {32} \right)^x} \cr & {\text{Then,}}{2^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {\left( {{2^2}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{3.6}} \times {2^{\left( {2 \times 3.6} \right)}} \times {2^{\left( {2 \times 3.6} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{\left( {3.6 + 7.2 + 7.2} \right)}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {2^{18}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{3.6}} \times {({2^5})^{2.3}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{\left( {3.6 + 2.3} \right)}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow {\left( {{2^5}} \right)^{5.9}} = {\left( {{2^5}} \right)^x} \cr & \Leftrightarrow x = 5.9 \cr} $$

$$\eqalign{ & {\text{Let }}{25^{2.7}} \times {5^{4.2}} \div {5^{5.4}} = {25^x} \cr & {\text{Then, }}{25^{2.7}} \times {5^{(4.2 - 5.4)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times {5^{( - 1.2)}} = {25^x} \cr & \Leftrightarrow {25^{2.7}} \times \frac{1}{{{5^{1.2}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{25}^{2.7}}}}{{{{\left( {{5^2}} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow \frac{{{{\left( {25} \right)}^{2.7}}}}{{{{\left( {25} \right)}^{0.6}}}} = {25^x} \cr & \Leftrightarrow {25^x} = {25^{\left( {2.7 - 0.6} \right)}} = {25^{2.1}} \cr & \Leftrightarrow x = 2.1 \cr} $$

$$\eqalign{ & {\text{Let }}{8^{2.4}} \times {2^{3.7}} \div {\left( {16} \right)^{1.3}} = {2^x} \cr & {\text{Then,}}{\left( {{2^3}} \right)^{2.4}} \times {2^{3.7}} \div {\left( {{2^4}} \right)^{1.3}} = {2^x} \cr & \Leftrightarrow {2^{\left( {3 \times 2.4} \right)}} \times {2^{3.7}} \div {2^{\left( {4 \times 1.3} \right)}} = {2^x} \cr & \Leftrightarrow {2^{7.2}} \times {2^{3.7}} \div {2^{5.2}} = {2^x} \cr & \Leftrightarrow {2^x} = {2^{\left( {7.2 + 3.7 - 5.2} \right)}} \cr & \Leftrightarrow {2^x} = {2^{5.7}} \cr & \Leftrightarrow x = 5.7 \cr} $$

$$\eqalign{ & {{\text{3}}^{x + y}}{\text{ = 81}} \cr & {{\text{3}}^{x + y}}{\text{ = }}{{\text{3}}^4} \cr & x + y = 4.....(i) \cr & {\text{8}}{{\text{1}}^{x - y}}{\text{ = 3}} \cr & {3^{4x - 4y}}{\text{ = }}{{\text{3}}^1} \cr & 4x - 4y{\text{ = 1}}....{\text{(ii)}} \cr & {\text{From equation (i) and (ii)}} \cr & 4x - 4y = 1 \cr & 4x + 4y = 16 \cr & 8x = 17 \cr & x = \frac{{17}}{8} \cr} $$

$$\eqalign{ & {\left[ {{{\left( {\root 5 \of {{x^{ - \frac{3}{5}}}} } \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {{{\left( {{x^{ - \frac{3}{{25}}}}} \right)}^{ - \frac{5}{3}}}} \right]^{ - 5}} \cr & = {\left[ {\left( {{x^{\frac{1}{5}}}} \right)} \right]^{ - 5}} \cr & = {x^{ - \frac{1}{5} \times 5}} \cr & = {x^{ - 1}} \cr & = \frac{1}{x} \cr} $$

$$\eqalign{ & \root 4 \of {3x + 1} = 2 \cr & \Rightarrow {\left( {\root 4 \of {3x + 1} } \right)^4} = {2^4} \cr & \Rightarrow {\left( {3x + 1} \right)^{4 \times \frac{1}{4}}} = 16 \cr & \Rightarrow 3x = 15 \cr & \Rightarrow x = 5 \cr} $$