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$If 1 0^{x} = \frac{1}{2} then 1 0^{- 8 x} = ?$ ${\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?$

1¹/₂₅₆
216
380
4256

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} 1 0^{- 8 x} \\ = {[10^{x}]}^{- 8} \\ = {[\frac{1}{2}]}^{- 8} \\ = 2^{8} \\ = 256 \end{aligned}

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

${(48)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} = ?$ ${\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?$

1¹/₃
2¹/₄₈
31
448

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(48)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} \\ = {(16 \times 3)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} \\ = {(16)}^{- \frac{2}{7}} \times {(3)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} \\ = {(16)}^{(- \frac{2}{7} - \frac{5}{7})} \times {(3)}^{(- \frac{2}{7} - \frac{5}{7})} \\ = {(16)}^{(- \frac{7}{7})} \times {(3)}^{(- \frac{7}{7})} \\ = {(16)}^{- 1} \times {(3)}^{- 1} \\ = \frac{1}{16} \times \frac{1}{3} \\ = \frac{1}{48} \end{aligned}

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

$\begin{aligned} The value of \\ \frac{1}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{1}{{(32)}^{- \frac{1}{5}}} is = ? \end{aligned}$ $\eqalign{ & {\text{The value of}} \cr & \frac{1}{{{{\left( {216} \right)}^{ - \frac{2}{3}}}}}{\text{ + }}\frac{1}{{{{\left( {256} \right)}^{ - \frac{3}{4}}}}}{\text{ + }}\frac{1}{{{{\left( {32} \right)}^{ - \frac{1}{5}}}}}{\text{ is = ?}} \cr}$

1102
2105
3107
4109

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{{(216)}^{- \frac{2}{3}}} + \frac{1}{{(256)}^{- \frac{3}{4}}} + \frac{1}{{(32)}^{- \frac{1}{5}}} \\ = \frac{1}{{(6^{3})}^{- \frac{2}{3}}} + \frac{1}{{(4^{4})}^{- \frac{3}{4}}} + \frac{1}{{(2^{5})}^{- \frac{1}{5}}} \\ = \frac{1}{6^{3 \times \frac{(- 2)}{3}}} + \frac{1}{4^{4 \times \frac{(- 3)}{4}}} + \frac{1}{2^{5 \times \frac{(- 1)}{5}}} \\ = \frac{1}{6^{- 2}} + \frac{1}{4^{- 3}} + \frac{1}{2^{- 1}} \\ = (6^{2} + 4^{3} + 2^{1}) \\ = (36 + 64 + 2) \\ = 102 \end{aligned}

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

$\begin{aligned} [8 - {(\frac{4^{\frac{9}{4}} \sqrt{{2.2}^{2}}}{2 \sqrt{2^{- 2}}})}^{^{\frac{1}{2}}}] \\ is equal to = ? \end{aligned}$ $\eqalign{ & \left[ {8 - {{\left( {\frac{{{4^{\frac{9}{4}}}\sqrt {{{2.2}^2}} }}{{2\sqrt {{2^{ - 2}}} }}} \right)}^{^{\frac{1}{2}}}}} \right] \cr & {\text{is equal to = ?}} \cr}$

132
28
31
40

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} [8 - {(\frac{4^{\frac{9}{4}} \sqrt{{2.2}^{2}}}{2 \sqrt{2^{- 2}}})}^{^{\frac{1}{2}}}] \\ = [8 - {(\frac{2^{2 \times \frac{9}{4}} \sqrt{2^{1 + 2}}}{2 \sqrt{\frac{1}{4}}})}^{^{\frac{1}{2}}}] \\ = [8 - {(\frac{2^{\frac{9}{2}} {.2}^{\frac{3}{2}}}{2 \times \frac{1}{2}})}^{^{\frac{1}{2}}}] \\ = [8 - {(2^{\frac{12}{2}})}^{^{\frac{1}{2}}}] \\ = [8 - (2^{6 \times \frac{1}{2}})] \\ = [8 - (2^{3})] \\ = [8 - 8] \\ = 0 \end{aligned}

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

$\begin{aligned} \frac{12}{3 + \sqrt{5} + 2 \sqrt{2}} \\ is equal to = ? \end{aligned}$ $\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & {\text{is equal to = ?}} \cr}$

11 - √5 + √2 + √16
21 + √5 + √2 - √10
31 + √5 + √2 + √10
41 - √5 + √2 + √10

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{12}{3 + \sqrt{5} + 2 \sqrt{2}} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{[(3 + \sqrt{5}) + 2 \sqrt{2}] [(3 + \sqrt{5}) - 2 \sqrt{2}]} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{{(3 + \sqrt{5})}^{2} - {(2 \sqrt{2})}^{2}} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{9 + 5 + 6 \sqrt{5} - 8} \\ = \frac{12 (3 + \sqrt{5} - 2 \sqrt{2})}{6 \sqrt{5} + 6} \\ = \frac{2 (3 + \sqrt{5} - 2 \sqrt{2})}{\sqrt{5} + 1} \\ = \frac{2 (3 + \sqrt{5} - 2 \sqrt{2})}{(\sqrt{5} + 1) (\sqrt{5} - 1)} \\ = \frac{2 (3 \sqrt{5} + 5 - 2 \sqrt{10} - 3 - \sqrt{5} + 2 \sqrt{2})}{5 - 1} \\ = \frac{2 (2 \sqrt{5} + 2 \sqrt{2} - 2 \sqrt{10} + 2)}{4} \\ = \frac{2 \times 2 (\sqrt{5} + \sqrt{2} - \sqrt{10} + 1)}{4} \\ = \sqrt{5} + \sqrt{2} - \sqrt{10} + 1 \end{aligned}

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

$\begin{aligned} The value of \\ 5^{\frac{1}{4}} \times {(125)}^{0.25} is = ? \end{aligned}$ $\eqalign{ & {\text{The value of}} \cr & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}{\text{ is = ?}} \cr}$

1√5
25
35√5
425

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} 5^{\frac{1}{4}} \times {(125)}^{0.25} \\ = 5^{0.25} \times {(5^{3})}^{0.25} \\ = 5^{0.25} \times 5^{(3 \times 0.25)} \\ = 5^{0.25} \times 5^{0.75} \\ = 5^{(0.25 + 0.75)} \\ = 5^{1} \\ = 5 \end{aligned}

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

What will come in place of both the question marks in the following question $\frac{{(?)}^{\frac{2}{3}}}{42} = \frac{5}{{(?)}^{\frac{1}{3}}}$ $\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}$

110
210√2
3√20
4210

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} Let \frac{{(x)}^{\frac{2}{3}}}{42} = \frac{5}{{(x)}^{\frac{1}{3}}} \\ Then, \\ x^{\frac{2}{3}} . x^{\frac{1}{3}} = 42 \times 5 \\ = 210 \\ \Rightarrow x^{(\frac{2}{3} + \frac{1}{3})} = 210 \\ \Rightarrow x = 210 \end{aligned}

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

$\begin{aligned} If x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} and \\ y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} then \\ (x + y) equals ? \end{aligned}$ $\eqalign{ & {\text{If }}x = \frac{{\sqrt 5 + \sqrt 3 }}{{\sqrt 5 - \sqrt 3 }}{\text{ and}} \cr & y = \frac{{\sqrt 5 - \sqrt 3 }}{{\sqrt 5 + \sqrt 3 }}{\text{ then}} \cr & \left( {x + y} \right){\text{equals ?}} \cr}$

18
216
32√15
42(√5 + √3)

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \\ \Rightarrow x = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} \\ \Rightarrow x = \frac{{(\sqrt{5} + \sqrt{3})}^{2}}{2} \\ Similarly \\ y = \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}} \\ \Rightarrow y = \frac{{(\sqrt{5} - \sqrt{3})}^{2}}{2} \\ Now, x + y \\ = \frac{5 + 3 + 2 \sqrt{15} + 5 + 3 - 2 \sqrt{15}}{2} \\ = \frac{16}{2} \\ = 8 \end{aligned}

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

$\begin{aligned} [\frac{{(0.73)}^{3} + {(0.27)}^{3}}{{(0.73)}^{2} + {(0.27)}^{2} - (0.73) \times (0.27)}] \\ simplifies to ? \end{aligned}$ $\eqalign{ & \left[ {\frac{{{{\left( {0.73} \right)}^3} + {{\left( {0.27} \right)}^3}}}{{{{\left( {0.73} \right)}^2} + {{\left( {0.27} \right)}^2} - \left( {0.73} \right) \times \left( {0.27} \right)}}} \right] \cr & {\text{simplifies to ?}} \cr}$

11
20.4087
30.73
40.27

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} let \\ a = 0.73 \\ b = 0.27 \\ = \frac{a^{3} + b^{3}}{a^{2} + b^{2} - a b} \\ = \frac{(a + b) (a^{2} + b^{2} - a b)}{(a^{2} + b^{2} - a b)} \\ = (a + b) \\ = (0.73 + 0.27) \\ = 1 \end{aligned}

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

$\frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = ?$ $\frac{1}{{\sqrt 3 + \sqrt 4 }} + \frac{1}{{\sqrt 4 + \sqrt 5 }} + \frac{1}{{\sqrt 5 + \sqrt 6 }} + \frac{1}{{\sqrt 6 + \sqrt 7 }} + \frac{1}{{\sqrt 7 + \sqrt 8 }} + \frac{1}{{\sqrt 8 + \sqrt 9 }} = ?$

1√3
23√3
33 - √3
45 - √3

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Answer:- 1

Explanation:-

Solution:

\begin{aligned} \frac{1}{\sqrt{3} + \sqrt{4}} \\ = \frac{1}{\sqrt{4} + \sqrt{3}} \times \frac{\sqrt{4} - \sqrt{3}}{\sqrt{4} - \sqrt{3}} \\ = \frac{\sqrt{4} - \sqrt{3}}{1} \\ = \sqrt{4} - \sqrt{3} \\ Similarly \\ \frac{1}{\sqrt{4} + \sqrt{5}} = \sqrt{5} - \sqrt{4} \\ \Rightarrow \frac{1}{\sqrt{5} + \sqrt{6}} = \sqrt{6} - \sqrt{5} \\ \Rightarrow \frac{1}{\sqrt{6} + \sqrt{7}} = \sqrt{7} - \sqrt{6} \\ \Rightarrow \frac{1}{\sqrt{7} + \sqrt{8}} = \sqrt{8} - \sqrt{7} \\ \Rightarrow \frac{1}{\sqrt{8} + \sqrt{9}} = \sqrt{9} - \sqrt{8} \\ Now put values \\ \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} \\ \Rightarrow \sqrt{9} - \sqrt{3} \\ \Rightarrow 3 - \sqrt{3} \end{aligned}

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

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Surds and Indices - 03

$If 1 0^{x} = \frac{1}{2} then 1 0^{- 8 x} = ?$ ${\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?$

${(48)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} = ?$ ${\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?$

$\begin{aligned} \frac{12}{3 + \sqrt{5} + 2 \sqrt{2}} \\ is equal to = ? \end{aligned}$ $\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & {\text{is equal to = ?}} \cr}$

$\begin{aligned} The value of \\ 5^{\frac{1}{4}} \times {(125)}^{0.25} is = ? \end{aligned}$ $\eqalign{ & {\text{The value of}} \cr & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}{\text{ is = ?}} \cr}$

What will come in place of both the question marks in the following question $\frac{{(?)}^{\frac{2}{3}}}{42} = \frac{5}{{(?)}^{\frac{1}{3}}}$ $\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}$

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Surds and Indices - 03

01. If 10x = 12 then 10−8x=?If 10x = 12 then 10−8x=?{\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?

02. (48)−27×(16)−57×(3)−57=?(48)−27×(16)−57×(3)−57=?{\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?

05. 123+5–√+22–√is equal to = ?123+5+22is equal to = ?\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & {\text{is equal to = ?}} \cr}

06. The value of514×(125)0.25 is = ?The value of514×(125)0.25 is = ?\eqalign{ & {\text{The value of}} \cr & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}{\text{ is = ?}} \cr}

07. What will come in place of both the question marks in the following question (?)2342=5(?)13(?)2342=5(?)13\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}

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$If 1 0^{x} = \frac{1}{2} then 1 0^{- 8 x} = ?$ ${\text{If 1}}{{\text{0}}^x}{\text{ = }}\frac{1}{2}{\text{ then 1}}{{\text{0}}^{ - 8x}} = ?$

${(48)}^{- \frac{2}{7}} \times {(16)}^{- \frac{5}{7}} \times {(3)}^{- \frac{5}{7}} = ?$ ${\left( {48} \right)^{ - \frac{2}{7}}} \times {\left( {16} \right)^{ - \frac{5}{7}}} \times {\left( 3 \right)^{ - \frac{5}{7}}} = ?$

$\begin{aligned} \frac{12}{3 + \sqrt{5} + 2 \sqrt{2}} \\ is equal to = ? \end{aligned}$ $\eqalign{ & \frac{{12}}{{3 + \sqrt 5 + 2\sqrt 2 }} \cr & {\text{is equal to = ?}} \cr}$

$\begin{aligned} The value of \\ 5^{\frac{1}{4}} \times {(125)}^{0.25} is = ? \end{aligned}$ $\eqalign{ & {\text{The value of}} \cr & {{\text{5}}^{\frac{1}{4}}} \times {\left( {125} \right)^{0.25}}{\text{ is = ?}} \cr}$

What will come in place of both the question marks in the following question $\frac{{(?)}^{\frac{2}{3}}}{42} = \frac{5}{{(?)}^{\frac{1}{3}}}$ $\frac{{{{\left( ? \right)}^{\frac{2}{3}}}}}{{42}} = \frac{5}{{{{\left( ? \right)}^{\frac{1}{3}}}}}$

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