Given log₈ x + log₄ x + log₂ x = 11
⇒ \(\frac { 1 }{ { log }_{ x }^{ 8 } } +\frac { 1 }{ { log }_{ x }^{ 4 } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \) = 11
⇒ \(1-\frac { 1 }{ { { log }_{ x }^{ { 2 }^{ 3 } } } } +\frac { 1 }{ { log }_{ x }^{ { 2 }^{ 2 } } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \)  = 11
⇒  \(\frac { 1 }{ 3{ { log }_{ x }^{ 2 } } } +\frac { 4 }{ 2{ { log }_{ x }^{ 2 } } } +\frac { 1 }{ { log }_{ x }^{ 2 } } \) = 11
⇒ \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 1 }{ 3 } +\frac { 1 }{ 2 } +1 \right) \)  = 11
⇒   \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 2+3+6 }{ 6 } \right) \)= 11
⇒ \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \left( \frac { 11 }{ 6 } \right) \) = 11
⇒   \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } =11\times \frac { 6 }{ 11 } =6\)
⇒  \(\frac { 1 }{ { { log }_{ x }^{ 2 } } } \) = 6
⇒  log₂^x = 6 
⇒ 2⁶ = x
⇒ x = 64
∴ x = 64

Given \({ log }_{ 9 }^{ 27 }-{ log }_{ 27 }^{ 9 }\)
= \({ log }_{ 9 }^{ 3^3 }-{ log }_{ 27 }^{ 3^2 }\)
=  3 log₉3-2 log₂₇3  [By power rule]
=   \(\frac { 3 }{ { log }_{ 3 }9 } -\frac { 2 }{ { log }_{ 3 }27 } \)    [By change of the base rule]
= \(\frac { 3 }{ { log }_{ 3 }{ 3 }^{ 2 } } -\frac { 2 }{ { log }_{ 3 }{ 3 }^{ 3 } } \)
= \(\frac { 3 }{ { 2\quad log }_{ 3 }{ 3 } } -\frac { 2 }{ { 3\quad log }_{ 3 }{ 3 } } \)
=  \(\frac { 3 }{ 2 } -\frac { 2 }{ 3 } \)      [∵ log₃³ = 1]
= \(\frac { 9-4 }{ 6 } \)

The means \(-{{1}\over{4}}-3\le{3\over4}x\le{1\over 4}.\)
\(⇒-{{1}\over{4}}-3≤-{3\over4}x≤{1\over4}-3\)
\(⇒-{13\over 4}≤-{3\over4}x≤-{11\over4}\)
Multiplying by 4 throughout we get,
\(-13≤-3x≤-11\)
\({-13\over -3}≥x≥{-11\over3}.\)
\(\therefore\) The Solution set is \(\left[ \frac { -11 }{ 3 } ,\frac { 13 }{ 3 } \right] \)

\(\frac { 7+\sqrt { 6 } }{ 3-\sqrt { 2 } } \times \frac { 3+\sqrt { 2 } }{ 3+\sqrt { 2 } } \)
⇒  \(\frac { (7+\sqrt { 6 } )(3+\sqrt { 2 } ) }{ { 3 }^{ 2 }-(\sqrt { 2 } )^{ 2 } }\)
\( \Rightarrow \frac { 21+7\sqrt { 2 } +3\sqrt { 6 } +\sqrt { 12 } 9-2 }{ 9-2 }\)
\( \Rightarrow \frac { 21+7\sqrt { 2 } +3\sqrt { 6 } +\sqrt { 3 } }{ 7 } \)

Given number is \(\frac { 1 }{ { 2 }^{ 1000 } } \)
we know 1000 < 1001
\(\Rightarrow\) 2¹⁰⁰⁰ < 2¹⁰⁰¹
\(\Rightarrow\) \(\frac { 1 }{ { 2 }^{ 1000 } } <\frac { 1 }{ { 2 }^{ 1001 } } \)
\(\therefore\)  A positive number smaller than \(\frac { 1 }{ { 2 }^{ 1000 } } is\frac { 1 }{ { 2 }^{ 1001 } } \)

Let the two irrational numbers be \(5+\sqrt { 7 } \) and \(7-\sqrt { 7 } \)
Their sum = \(\left( 5+\sqrt { 7 } \right) +\left( 7-\sqrt { 7 } \right) \ = \ 5+\sqrt { 7 } +7-\sqrt { 7 } \)
= 5 + 7 = 12 which is a rational number.
Consider  the two irrational numbers \(4+\sqrt { 6 } \) and \(4-\sqrt { 6 } \)
Their product = \(\left( 4+\sqrt { 6 } \right) +\left( 4-\sqrt { 6 } \right) ={ 4 }^{ 2 }-{ \left( \sqrt { 6 } \right) }^{ 2 }\)= 16 - 6 = 10 which is a rational number.

Suppose √3 is a rational number
Then √3 can be written as \(√3={m\over n}\)
Where m and n are rational numbers with no common factors other than 1.
Squaring both sides we get,
\(3={m^2\over n^2}⇒3n^2=m^2\)
multiplying by 2 we get
6n² = 2m² ⇒ 3(2n²) = 2 m²
Since 2n² is divisible by 2, m² is also an even number
⇒ m must be even
⇒ m = 2k for some natural number k
⇒ 3n² = (2k)²
⇒ 3n² = 4k²
⇒ n is also an even number
Thus both m and n are even numbers having a common factor.
This contradicts our initial assumption that m and n do not have a common factor
Hence √3 cannot be a rational number.
⇒ √3 is an irrational number.
Hence proved.

Given \(\left( { x }^{ +\frac { 1 }{ 2 } }+{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ 2 } +{ 2x }^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ -\frac { 1 }{ 2 } } } =\frac { 9 }{ 2 } \)
⇒  \(x+\frac { 1 }{ x } +2=\frac { 9 }{ 2 } \)
⇒ \(x+\frac { 1 }{ x } =\frac { 9 }{ 2 } -2=\frac { 9-4 }{ 2 } =\frac { 5 }{ 2 } \)
Consider \(\left( { x }^{ \frac { 1 }{ 2 } }-{ x }^{ -\frac { 1 }{ 2 } } \right) ^{ 2 }\)
= \(x+\frac { 1 }{ x } -2x^{ \frac { 1 }{ 2 } }.\frac { 1 }{ { x }^{ \frac { 1 }{ 2 } } } \)
= \(x+\frac { 1 }{ x } -2\)    [From(1)]
=   \(\frac { 5 }{ 2 } -2=\frac { 5-4 }{ 2 } =\frac { 1 }{ 2 } \)       [using 1]
∴  \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\pm \frac { 1 }{ \sqrt { 2 } } \)
⇒ \(x^{ \frac { 1 }{ 2 } }-x^{ -\frac { 1 }{ 2 } }=\frac { 1 }{ \sqrt { 2 } } \)   since x > 1

Let the two distinct irrational numbers be \(\left( 2+\sqrt { 3 } \right) \) and \(\left( 4+\sqrt { 3 } \right) \).
Their difference is \(\left( 2+\sqrt { 3 } \right) -\left( 4+\sqrt { 3 } \right) =2+\sqrt { 3 } -4-\sqrt { 3 } =2-4=-2\) which is rational.

Since \(\sqrt { 7 } \) is an irrational number,  \(\sqrt { 7 } \) \(\in\) R.
Since \(\frac { -1 }{ 4 } \) is a negative rational number \(\frac { -1 }{ 4 } \) \(\in\) Q
0 is an integer and 0 \(\in\)Z.
3.14 = \(\pi \) is a non-recurring and non-terminating decimal.
\(\therefore\) 3.14 is an irrational number \(\Rightarrow\) 3.14 \(\in\) R-Q
4 is a positive integer \(\Rightarrow\) 4 \(\in\)R-Q.
\(\frac { 22 }{ 7 } \) = 3.14 \(\in\) R. Which is an irrational number.