i) 64 = 2x
We know that
64 = 2 × 32
= 2 × 2 × 16
= 2 × 2 × 2 × 8
= 2 × 2 × 2 × 2 × 4
= 2 × 2 × 2 × 2 × 2 × 2
64 = 26
⇒ x = 6
ii) 100 = 5b
Here also 100 cannot be written as any power of 5.
i.e., there exists no integer for b such that 5b = 100
iii) \(\frac{1}{81}=3^c\)
We know that 81 = 3 x 27
= 3 × 3 × 9
= 3 × 3 × 3 × 3
= 34
∴ \(\frac{1}{81}=3^{-4}\) [∵a-3 = \(\frac{1}{a^m}\)]
∴ c = – 4
iv) 100 = 102
z = 2
v) \(\frac{1}{256}\)= 4a
We know that 256 = 4 × 64
= 4 × 4 × 16
= 4 × 4 × 4 × 4
∴ \(\frac{1}{256}\) = 4
∴ a = – 4