If log 2 = 0.30103, the number of digits in 2⁶⁴ is:

(b) log (2 + 3) = log 5 and log (2 x 3) = log 6 = log 2 + log 3

     ∴ log (2 + 3) ≠ log (2 x 3)

(c) Since log_a 1 = 0, so log₁₀ 1 = 0.

(d) log (1 + 2 + 3) = log 6 = log (1 x 2 x 3) = log 1 + log 2 + log 3.

So, (b) is incorrect.

⟹ log₁₀ 5 + log₁₀ (5x + 1) = log₁₀ (x + 5) + log₁₀ 10

⟹ log₁₀ [5 (5x + 1)] = log₁₀ [10(x + 5)]

⟹ 5(5x + 1) = 10(x + 5)

⟹ 5x + 1 = 2x + 10

⟹ 3x = 9

⟹ x = 3.

Then, 2ⁿ = 16 = 2⁴     ⟹     n = 4.

∴ log₂ 16 = 4.

∴ log_x y = 100

⟹ y = x¹⁰⁰

⟹ y = (2¹⁰)¹⁰⁰     [put value of x]

⟹ y = 2¹⁰⁰⁰.