\(\frac { 2\times { 3 }^{ 4 }\times { 2 }^{ 5 } }{ 9\times { 4 }^{ 2 } } =\frac { 2\times { 3 }^{ 4 }\times { 2 }^{ 5 } }{ { 3 }^{ 2 }\times { \left( { 2 }^{ 2 } \right) }^{ 2 } } =\frac { 2\times { 2 }^{ 5 }\times { 3 }^{ 4 } }{ { 3 }^{ 2 }\times { 2 }^{ 2\times 2 } } =\frac { { 2 }^{ 1+5 }\times { 3 }^{ 4 } }{ { 2 }^{ 4 }\times { 3 }^{ 2 } } =\frac { { 2 }^{ 6 }\times { 3 }^{ 4 } }{ { 2 }^{ 4 }\times { 3 }^{ 2 } } \)
\(={ 2 }^{ 6-4 }\times { 3 }^{ 4-2 }={ 2 }^{ 2 }\times { 3 }^{ 2 }\)=4x9=36

2⁴=4²

2¹⁰

225 = 3²x5²

We have, 2⁴ x 3² = 2 x 2 x 2 x 2 x3 x3 = 16 x 9 = 144
Hence, the value of 2⁴ x 3² is 144.

Wehave,0x10² = 0x(10x10)=0x100=0
Hence, the value of 0 x10² is 0.

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]

We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

Here, base = 2 and exponent = 7, since 2 repeated 7 times.

We have, 729 = 3 x 3 x 3 x 3 x 3 x 3 = 3⁶

Here, base= 3 and exponent = 6, since 3 repeated 6 times.

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

561

Given, 2ⁿ⁺²- 2ⁿ = c x 2ⁿ
⇒ 2ⁿ⁺² = 2ⁿ x 2², 2ⁿ⁺¹ = 2ⁿ x 2¹
So, 2ⁿ x 2² - 2ⁿ x 2¹ + 2ⁿ = c x 2ⁿ
Taking ~ common from both the side, we get
2ⁿ (2²-2¹)= c x 2ⁿ
⇒ (2²-2¹+1)c
∴ c = 4-2+1   [∵ 2²= 2 x 2=4, 2¹=2]
=2 +1 =3

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, 8.01 x 10⁷
∵ 10⁷ =10000000
and 8.01 = 801 x10^-2
So, 8.01 x 10⁷ = 801 x10^-2 x 10⁷ = 801 x 10⁵
= 801 x 100000 [∵ 10⁵ =100000]
= 80100000

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, \(\frac { p }{ q } =\left( \frac { 3 }{ 2 } \right) ^{ 2 }\div \left( \frac { 9 }{ 4 } \right) ^{ 0 }\)
∵ \(\left( \frac { 9 }{ 4 } \right) ^{ 0 }=\{ \left( \frac { 3 }{ 2 } \right) ^{ 2 }\} ^{ 0 }=\left( \frac { 3 }{ 2 } \right) ^{ 2\times 0 }=\left( \frac { 3 }{ 2 } \right) ^{ 0 }\)
Now, \(\left( \frac { 3 }{ 2 } \right) ^{ 3 }\div \left( \frac { 3 }{ 2 } \right) ^{ 0 }\)
Let a= \(\frac { 3 }{ 2 } \)
So, \(\frac { p }{ q } =\left( \frac { 3 }{ 2 } \right) ^{ 2-0 }=\left( \frac { 3 }{ 2 } \right) ^{ 2\times 3 }\)    [∵ a^m ÷ aⁿ =a^m-n]
Now, \(\left( \frac { p }{ q } \right) ^{ 3 }=\{ \left( \frac { 3 }{ 2 } \right) ^{ 2 }\} ^{ 3 }=\left( \frac { 3 }{ 2 } \right) ^{ 2x3 }=\left( \frac { 3 }{ 2 } \right) ^{ 6 }\)
= \(\frac { 3\times 3\times 3\times 3\times 3\times 3 }{ 2\times 2\times 2\times 2\times 2\times 2 } =\frac { 729 }{ 64 } \).

Given, \(\left( \frac { 2 }{ 9 } \right) ^{ 3 }\times \left( \frac { 2 }{ 9 } \right) ^{ 6 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
We know that. a^m x aⁿ = a^m+n
Let a=\(\frac { 2 }{ 9 } \)
So, \(\left( \frac { 2 }{ 9 } \right) ^{ 3 }\times \left( \frac { 2 }{ 9 } \right) ^{ 6 }=\left( \frac { 2 }{ 9 } \right) ^{ 3+6 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
⇒ \(\left( \frac { 2 }{ 9 } \right) ^{ 9 }=\left( \frac { 2 }{ 9 } \right) ^{ 2m-1 }\)
If am =an, then m = n
So, 9=2m-1 ⇒ 9+1 = 2m
∴ m=\(\frac { 10 }{ 2 } \)=5

Given, \(\frac { p }{ q } ={ \left( \frac { 3 }{ 4 } \right) }^{ 18 }\div { \left( \frac { 3 }{ 4 } \right) }^{ 17 }\)
\(\because\) a^m \(\div\)aⁿ =a^{m-n
Let \(a=\frac { 3 }{ 4 } \)
So, \(\frac { p }{ q } ={ \left( \frac { 3 }{ 4 } \right) }^{ 18-17 }={ \left( \frac { 3 }{ 4 } \right) }^{ 1 }=\frac { 3 }{ 4 } \)
\(\therefore\) \({ \left( \frac { p }{ q } \right) }^{ 3 }={ \left( \frac { 3 }{ 4 } \right) }^{ 3 }=\frac { 3\times 3\times 3 }{ 4\times 4\times 4 } =\frac { 27 }{ 64 } \).}

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 \(\frac { 4 }{ 9 } \)

(a) \(\frac{12^4\times 9^3\times 4}{6^3\times 8^2\times 27} =\frac{(3\times 2^2)^4\times (3^2)^3\times 2^2}{(2\times 3)^3\times (2^3)^2\times 3^3}\)
\(=\frac{3^4\times 2^8\times 3^6\times 2^2}{2^3\times3^3\times2^6\times3^3}\)
\(=\frac{2^{8+2}\times 3^{6+4}}{2^{6+3}\times 3^{3+3}}\)
\(=\frac{2^{10}\times 3^{10}}{2^9\times 3^6}=2^{10-9}\times 3^{10-6}\)
= 2 x 3⁴ = 2 x 81 = 162.
(b) 2³ x a³ x 5a⁴
= 8 x a³ x 5 x a⁴
= 8 x 5 x a³ x a⁴
= 40 x a³ + 4
= 40 x a⁷
= 40a⁷.

\(⇒\ \left(1\over 5\right)^{5+19}=\left(1\over 5\right)^{8x}\)
[∵ a^m x aⁿ = a^m+n]
\(⇒\ \left(1\over 5\right)^{24}=\left(1\over 5\right)^{8x}\)
When bases are equal, then by equating their exponents, we get
8x = 24
\(∴\ x={24\over 8}=3\)

\((a)\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\times\left(-{4\over 3}\right)\)
\(={(-4)\times(-4)\times(-4)\times(-4)\times(-4)\over 3\times3\times3\times3\times3}\)
\(={(-4)^5\over (3)^5}=\left(-4\over 3\right)^5\)
\((b)\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\times\left(-{8\over 3}\right)\)
\(={(-8)\times(-8)\times(-8)\times(-8)\times(-8)\over 3\times3\times3\times3\times3}\)
\(={(-8)^5\over (3)^5}=\left(-8\over 3\right)^5\)

(25)^n-1 + 100 = 5^(2n-1)
⇒ (5²)^n-1 + 100 = 5^(2n-1)
⇒ 5^2n-2 + 100 = 5^2n-1
⇒ 5^2n-2 - 5^2n-1 = - 100
⇒ 5^{2n - 1} - 5^{2n- 2} = 100
⇒ 5^2n-2 x (5 -1) = 100
⇒ 5^2n-2 x 4 = 100
\(⇒\ 5^{2n-2}={100\over 4}=25\)
Thus, 5^{2n- 2} = 5²
As base is same on both the sides
∴ 2n-2 = 2
⇒ 2n=2+2
⇒ 2n = 4
\(⇒\ n={4\over 2}=2\)

\(\frac{9^n\times 3^2\times 3^n-(27)^2}{(3^3)^5\times 2^3}=\frac{1}{27}\)
\(\Rightarrow \frac{(3\times 3)^n\times 3^2\times3^n-(3\times3\times3)^n}{3^{15}\times 2^3}=\frac{1}{3\times3\times3}\)
\(\Rightarrow \frac{(3^2)^n\times3^{n+2}-(3^3)^n}{3^{15}\times2^3}=\frac{1}{3^3}\)
\(\Rightarrow \frac{3^{2n}\times 3^{n+2}-3^{2n}}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{2n+n+2}-3^{3n}}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{2n+n+2}-3^{3n}}{3^{15}\times2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{3n}(3^2-1)}{3^{15}\times 2^3}=3^{-3}\)
\(\Rightarrow \frac{3^{3n-15}(9-1)}{2^3}=3^{-3}\)
\(\Rightarrow 3^{3n-15}\times \frac{8}{8}=3^{-3}\)
\(\Rightarrow 3^{3n-15}=3^{-3}\)
As base are same on both sides, so
3n - 15 = -3
\(\Rightarrow \) 3n = -3 + 15
\(\Rightarrow \) 3n = 12
thus,  \(n=\frac{12}{3}=4\)