∵ x=(100)^1-4 ÷ (100)⁰=(100)^-3    [∵ (100)⁰=1]
∴ \(x^{-3}=\frac{1}{x^{3}}=\frac { 1 }{ \left\{ { \left( 100 \right) }^{ -3 } \right\} ^{ 3 } } =\frac{1}{100^{-9}}\)      [∵ (a^m)ⁿ=a^mn]
= 100⁹

We have, 5.8 x 10¹² = 5.8 x 1000000000000
= 5800000000000

3²

36025.248= \(3\times10^{4}+6\times10^{3}+2\times10^{1}+5\times10^{0}+2\times10^{-1}+4\times10^{-2}+8\times10^{-3}\)

We have , {\((\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}\)}^-1
\(\left\{ \frac { { \left( 1 \right) }^{ -1 } }{ { \left( 3 \right) }^{ -1 } } -\frac { { \left( 1 \right) }^{ -1 } }{ { \left( 4 \right) }^{ -1 } } \right\} ^{ -1 }\quad \quad \quad \left[ \because \quad \left( \frac { a }{ b } \right) ^{ m }=\frac { { a }^{ m } }{ { b }^{ m } } \quad \right] \)
\(=\left\{ \frac { 3 }{ 1 } -\frac { 4 }{ 1 } \right\} ^{ -1 }\quad \quad \quad \left[ \because \quad { a }^{ -m }=\frac { 1 }{ { a }^{ m } } \right] \)
= (3-4)^-1 = (-1)^-1=\(\frac{1}{(-1)^{1}}\)     \( \left[ \because \quad { a }^{ -m }=\frac { 1 }{ { a }^{ m } } \right] \)
=\(\frac{1}{-1}=-1\)

We have, 1256.249
=1\(\times\)1000+2\(\times\)100+5\(\times\)10+6\(\times\)1+\(\frac{2}{10}+\frac{4}{100}+\frac{9}{1000}\)
\(=1\times 10^{3}+2\times 10^{2}+5\times 10^{1}+6\times 10^{0}+2\times 10^{-1}+4\times 10^{-2}+9\times 10^{-3}\)

Multiplicative inverse of 10^-100 is 10¹⁰⁰.

Multiplicative inverse of 5^-3 is 5³.

We have,10^-5=\(\frac{1}{10^{5}}\)
Reciprocal of \(\frac{1}{10^{5}}\)=10⁵
∴ Multiplicative inverse of 10^-5 is 10⁵.    [\(10^{-5}\times 10^{5}=10^{0}=1\)]

Let x be the required number.
Then, \(\frac{(-15)^{-1}}{x}=(-15)^{-1}\)
=\(\frac{1}{(-15)x}=\frac{1}{(-15)}\)                  [∵ \((a)^{-1}=\frac{1}{a}\)]
⇒ \(\frac{1}{x}=1 ⇒ x=1\)
So, (-15)^-1 should be divided by 1 to get quotient (-15)^-1.

\((\frac{1}{2})^{-2}\div(\frac{1}{2})^{-3}\)=\((\frac{1}{2})^{-2}\times(\frac{1}{2})^{-3}\)
= \((\frac{1}{2})^{-2+3}=(\frac{1}{2})^{1}=\frac{1}{2}\)

0.00015

(i) We have, a^b + b^a = (-1)² + (2)^-1= 1+\(\frac{1}{2}\)
=\(\frac{2+1}{2}=\frac{3}{2}\)
(ii) We have, a^b - b^a=(-1)²-(2)^-1            [∵ aⁿ=\(\frac{1}{a^{n}}\)]
=1-\(\frac{1}{2}\)=\(\frac{2-1}{2}=\frac{1}{2}\)
(iii) We have,a^b x b^a=(-1)²x(2)^-1=1x\(\frac{1}{2}\)=\(\frac{1}{2}\)    [∵ aⁿ=\(\frac{1}{a^{n}}\)]
(iv) We have, a^b ÷ b^a=(-1)÷(2)^-1=\((-1)^{2}\times \frac{1}{(2)^{-1}}\)
=1x2=2     [∵ \(\frac{1}{a^{-n}}\)=aⁿ]

4.42 X10²