We have, 5³ or 3⁵
\(\therefore\) 5³ = 5 x 5 x 5 = 125 [multiply 5 three times]
and 3⁵ =3 x3 x3 x3 x3= 243 [multiply 3 five times]
\(\because\)243>125 \(\Rightarrow\) 3⁵ > 5³
Hence, 3⁵ is greater than 5³

We have, 4³ or 3⁴ \(\Rightarrow\) 4³ = 4 x 4 x 4 = 64
[here, multiply base = 4 three times, since exponent = 3]
and 3⁴ = 3 x 3 x 3 x 3 = 81
[here, multiply base = 3 four times, since exponent = 4]
\(\because\) 81 > 64 \(\Rightarrow\) 3⁴ > 4³
Hence, 3⁴ is greater than 4³.

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]

b x b x b x b = b⁴ [since, b is multiplied 4 times]

t x t = t² [since, t is multiplied 2 times]

11² = 11 x 11 = 121
Hence, the value of 11² is 121.

6234269

8.054 x 10⁹

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, 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.

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

(a) 1 light year = 9.460,000,000,000 km
= 946 x 10¹⁰km = \(\frac { 946 }{ 100 } \)x 10¹⁰ X 100 km
= 9,46 x 10¹²km.
(b) Average distance between Earth and Sun =1.496 x10⁸ km
\(\therefore\) Distance between Earth and Sun = \(\frac { 1.496 }{ 10000 } \times { 10 }^{ 8 }\times { 10 }^{ 4 }\)km
= 0,0001496 x 10¹²km
\(\Rightarrow\)9.46 > 0.0001496
So, the distance between Earth and Sun less than one light year.