(c) an irrational number
Let us assume log27 be a rational number. Then,
log27 = \(\frac{p}{q}\), where p,q ∈ I and q ≠ 0
⇒ \(2^{\frac{p}{q}}\) = 7 ⇒ 2p = 7q
This is not true as 2 is even and 7 is odd.
∴ Hence our assumption that log27 is a rational number is wrong.
∴ log27 is an irrational number.