\(-\frac { 11 }{ 7 } \)

(- 3)⁴

We have, 3 x 4⁴ = 3 x 4 x 4 x 4 x 4 = 3 x 256 = 768
Hence, the value of 3 x 4⁴ is 768

Number 540 can be expressed as 2² x3³ x5, which is the required form.

We have, 2⁸ or 8²
\(\therefore\) 2⁸=2x2x2x2x2x2x2x2=256
and 8² = 8 X8 = 64
\(\because\) 256>64 \(\Rightarrow\) 2⁸ > 8²
Hence, 2⁸ is greater than 8².

We have, 343 = 7 x 7 x 7 = 7³ [since, 7 is multiplied 3 times]
Hence, the exponential form of 343 is 7³

a x a x a x c x c x c x c x d = a³ x c⁴ x d  [since, a is multiplied 3 times, c is multiplied 4 times and d is multiplied 1 time]

6 x 6 x 6 x 6 = 6⁴ [since, 6 is multiplied 4 times]

9³ = 9 x 9 x 9 = 729
Hence, the value of 9³ is 729.

We have, 343 = 7 x 7 x 7 = 7³

Here, base = 7 and exponent = 3, since 7 repeated 3 times.

We have [(3²)³ x 3⁶] \(\div\) 3⁶
= [3^2x3 X 3⁶] \(\div\) 3⁶              \([\because (a^m)^n =a^{mn}]\)
= [3⁶ X 3⁶]\(\div\) 3⁶
= [3^{6 + 6}] \(\div\) 3^{6                           \([\because a^m\times a^n =a^{m+n}]\)}
= 3¹² \(\div\) 3⁶
= 3^{12 - 6                                       \([\because a^m\div a^n =a^{m-n}]\)}
= 3⁶
Thus, [(3²)³ X 3⁶] \(\div\) 3⁶ = 3⁶.

we have \([{5^6\over 5^3}]\times 5^2\)  =[5^6-3]x5²      \((\because {a^m\over a^n}=a^{m-n})\)
= 5³ X 5² 
= 5^{3 + 2}                                         \((\because a^m \times a^n =a^{m+n})\)
= 5⁵
Thus,\([{5^6\over 5^3}]\times 5^2=5^5\)

6234269

8.054 x 10⁹

\(\frac { 9 }{ 32 } \)

169

Given, \({ 2 }^{ n-5 }\times { 6 }^{ 2n-4 }=\frac { 1 }{ 12^{ 4 }\times 2 } \)
∵ 6^2n-4 = (2 x 3)^2n-4 = 2^2n-4 x 3^2n-4
and 12⁴ = (3 x 4)⁴ =(3 x 2 x 2)⁴ = 3⁴ x 2⁴ x 2⁴
So, 2^n-5 x 2^2n-4 x 3^2n-4 =\(\frac { 1 }{ { 3 }^{ 4 }\times { 2 }^{ 4 }\times { 2 }^{ 4 }\times { 2 }^{ 1 } } \)
 ⇒ 2^n-5+2n-4 x 3^2n-4 =\(\frac { 1 }{ { 3 }^{ 4 }\times { 2 }^{ 4+4+1 } } \)
[∵ a^m x aⁿ =a^m+n]
⇒ 2^3n-9 x 3^2n-4 = \(\frac { 1 }{ { 3 }^{ 4 }\times { 2 }^{ 9 } } \)
⇒ 2^3n-9 x 3^2n-4 = 3^-4 x 2^-9
[∵ a^-m=\(\frac { 1 }{ { a }^{ m } } \)]
a^m =aⁿ ⇒  m=n
So,3n-9 = -9
3n = - 9 + 9 = 0 ⇒ n = 0

Given, \(\frac { p }{ q } ={ \left( \frac { -2 }{ 3 } \right) }^{ 9 }\div { \left( \frac { -2 }{ 3 } \right) }^{ 8 }\)
\(\because\) a^m \(\div\)aⁿ =a^m-n
Let \(a=\left( \frac { -2 }{ 3 } \right) \)
So, \(\frac { p }{ q } ={ \left( \frac { -2 }{ 3 } \right) }^{ 9-8 }={ \left( \frac { -2 }{ 3 } \right) }^{ 1 }=\frac { -2 }{ 3 } \)
\(\therefore\) \({ \left( \frac { p }{ q } \right) }^{ 2 }={ \left( \frac { -2 }{ 3 } \right) }^{ 2 }\)
\(=\frac { -2\times \left( -2 \right) }{ 3\times 3 } =\frac { 4 }{ 9 } \)
Hence, the value of \({ \left( \frac { p }{ q } \right) }^{ 2 }\) is \(\frac { 4 }{ 9 } \)

Given, 1024
∵ 1024 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2¹⁰

The exponent form of 1024 is 2¹⁰.

We have, 2.7 x 10¹²; 1.5 x 10⁸
2.7 x 10¹² = 2.7 x 10 x 10 x 10 x 10 x 10x 10⁷
=\(\frac { 27 }{ 10 } \) x 10x 10x 10x 10 x 10x 10⁷
= 270000 x 10⁷
and 1.5x10⁸ = \(\frac { 15 }{ 10 } \) x10x10⁷=15x10⁷
\(\because\) 270000 > 15
\(\therefore\)270000x10⁷>15x10⁷
or 2.7 x 10¹²> 1.5x 10⁸
Hence, 2.7 x 10¹²is greater than 1.5 x 10⁸.

(-4)² or 16

\(\frac { 8 }{ 3 } \)

Given, \(\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \left( \frac { 125 }{ 27 } \right) }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \left( \frac { 9 }{ 16 } \right) } \)
∵ \(\frac { 125 }{ 27 } =\frac { 5\times 5\times 5 }{ 3\times 3\times 3 } =\frac { 5^{ 3 } }{ 3^{ 3 } } \)
and \(\frac { 9 }{ 16 } =\frac { (-3)\times (-3) }{ 4\times 4 } =\frac { (-3)^{ 2 } }{ { 4 }^{ 2 } } \)
So, \(\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \frac { { 5 }^{ 3 } }{ { 3 }^{ 3 } } }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \frac { (-3)^{ 2 } }{ { 4 }^{ 2 } } } =\frac { \left( -\frac { 3 }{ 4 } \right) ^{ 4 }\times \left( \frac { 5 }{ 3 } \right) ^{ 3 } }{ \left( \frac { 5 }{ 3 } \right) ^{ 2 }\times \left( -\frac { 3 }{ 4 } \right) ^{ 2 } } \)      \(\left[ \because \frac { { a }^{ n } }{ b^{ n } } =\left( \frac { a }{ b } \right) ^{ n } \right] \)
= \(\left( \frac { -3 }{ 4 } \right) ^{ 4-2 }\times \left( \frac { 5 }{ 3 } \right) ^{ 3-2 }\)     [∵ a^m ÷ aⁿ = a^m-n]
=\(\left( \frac { -3 }{ 4 } \right) ^{ 2 }\times \left( \frac { 5 }{ 3 } \right) ^{ 1 }\)
= \(\frac { (-3)\times (-3) }{ 4\times 4 } \times \frac { 5 }{ 3 } =\frac { 9 }{ 16 } \times \frac { 5 }{ 3 } =\frac { 3\times 5 }{ 16 } =\frac { 15 }{ 16 } \).

\(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times (125)^{ 2/3 } }{ (27)^{ -2/3 }\times (32)^{ -1/5 } } \)
∵ 125 = (5)³ = 5 x 5 x 5
So, (125)^2/3 =(5)^{3 x 2/3} =5² and 27=(3)³
∴ (27)^-2/3 = {(3)³}^-2/3 =\({ (3 })^{ 3\times \frac { -2 }{ 3 } }\) =(3)²
32 = 2 x 2 x 2 x 2 x 2 = (2)⁵
So, (32)^-1/5 = {(2)⁵}^-1/5 = \({ (2) }^{ 5\times \frac { (-1) }{ 5 } }\)=(2)^-1
Now \(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times { 5 }^{ 2 } }{ (3)^{ -2 }\times (2)^{ -1 } } \)     \(\left[ \because a^{ -m }=\frac { 1 }{ { a }^{ m } } \right] \)
= 5^-2 x 3^-3 x 3² x 2¹ x 5²
=5^-2+2 x 3^-3+2 x 2¹
a^m x aⁿ =a^m+n
=5⁰ x 3^-1 x 2¹ = 1 x \(\frac { 1 }{ 3 } \times 2=\frac { 2 }{ 3 } \).

Given, 1.75 x1 0^-3
1.75 = 175 x 10^-2
So, 1.75 x 10^-3 =175 x 10^-3 x 10^-2
=175 x 10^-5          \(\left[ \because \frac { 1 }{ { 10 }^{ 5 } } ={ 10 }^{ -5 } \right] \)
= 0.00175