Given, 2⁵ x 4² x 2²
∵ 2⁵ = 2 x 2 x 2 x 2 x 2 = 32,
4² = 4 x 4 = 16 and 22 = 2 x 2 = 4
So, 32 x 16 x 4 = 2048

123870000000
For standard form, 123870000000 = 12387 x10000000 =12387 x 10⁷
\(\because\)12387 = 1.2387 x 10000 = 1.2387 x 10⁴
So, 12387x10⁷ = 1.2387 x 10⁴ x 10⁷ = 1.2387x 10¹¹

(- 3)²= 3²

2²

\(\frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 } =\frac { { \left( 2\times 3 \right) }^{ 4 }\times { \left( { 3 }^{ 2 } \right) }^{ 3 }\times { 2 }^{ 2 } }{ { \left( 2\times 3 \right) }^{ 3 }\times { \left( { 2 }^{ 3 } \right) }^{ 2 }\times { 3 }^{ 3 } } =\frac { { \left( { 2 }^{ 2 } \right) }^{ 4 }\times { \left( 3 \right) }^{ 4 }\times { 3 }^{ 2\times 3 }\times { 2 }^{ 2 } }{ { 2 }^{ 3 }\times { 3 }^{ 3 }\times { 2 }^{ 2\times 3 }\times { 3 }^{ 3 } } \)
=\(\frac { { 2 }^{ 8 }\times { 2 }^{ 2 }\times { 3 }^{ 4 }\times { 3 }^{ 6 } }{ { 2 }^{ 3 }\times { 2 }^{ 6 }\times { 3 }^{ 3 }\times { 3 }^{ 3 } } =\frac { { 2 }^{ 8+2 }\times { 3 }^{ 4+6 } }{ { 2 }^{ 3+6 }\times { 3 }^{ 3+3 } } =\frac { { 2 }^{ 10 }\times { 3 }^{ 10 } }{ { 2 }^{ 9 }\times { 3 }^{ 6 } } \)
\(={ 2 }^{ 10-9 }\times { 3 }^{ 10-6 }={ 2 }^{ 1 }\times { 3 }^{ 4 }=\)2x81 = 162
 

2744

3² x 5² x 7²

Base = 7, Exponent= 3

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

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

2⁶ = 2 x 2 x 2 x 2 x 2 x 2 = 64
Hence, the value of 2⁶ is 64

Given, 2¹⁹⁹⁸- 2¹⁹⁹⁷- 2¹⁹⁹⁶+ 2¹⁹⁹⁵= k.2¹⁹⁹⁵
\(\Rightarrow\)2¹⁹⁹⁵⁺³ - 2^{1995+ 2} - 2^{1995+ 1}+ 2¹⁹⁹⁵ x1 = k.2¹⁹⁹⁵
\(\Rightarrow\)2¹⁹⁹⁵ [2³ - 2² - 2¹ + 1] = k. 2¹⁹⁹⁵
\(\Rightarrow\)2¹⁹⁹⁵ [8 - 4 - 2 + 1] = k. 2¹⁹⁹⁵
\(\Rightarrow\)3=\(\frac { k.{ 2 }^{ 1995 } }{ { 2 }^{ 1995 } }\) \(\Rightarrow\) 3 = k
So, the value of k is 3.

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 \(\frac { 4 }{ 9 } \)

30⁶

(i) \(\frac { 1 }{ 256 } \)
(ii) \(-\frac { 8 }{ 27 } \)

Given, \({ \left( \frac { 1 }{ 5 } \right) }^{ 5 }\times { \left( \frac { 1 }{ 5 } \right) }^{ 19 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)
\(\Rightarrow ​​{ \left( \frac { 1 }{ 5 } \right) }^{ 5+19 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)      [\(\because\)a^m x aⁿ = a^m+n]
\(\Rightarrow ​​{ \left( \frac { 1 }{ 5 } \right) }^{ 24 }={ \left( \frac { 1 }{ 5 } \right) }^{ 8x }\)
Since, bases are equal, by equating their exponents, we get
8x = 24
\(\therefore\)  x = 24/8 = 3

\(\frac { { 5 }^{ -2 }\times { 3 }^{ -3 }\times (125)^{ 2/3 } }{ (27)^{ -2/3 }\times (32)^{ -1/5 } } \)
∵ 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 \(\frac { 1 }{ 3 } \times 2=\frac { 2 }{ 3 } \).