log_5(x) = 2 implies x = 5^2 = 25.

Since 49 = 7^2, log_7(49) = 2, thus x = 2.

From log_a(16) = 2, we have a^2 = 16, thus a = 4.

log_a(6) = log_a(2 * 3) = log_a(2) + log_a(3) = x + y.

log_a(6) = log_a(2 * 3) = log_a(2) + log_a(3) = x + y.

log_a(16) = log_a(4^2) = 2 * log_a(4) = 2 * 2 = 4.

log_a(4) = 2 implies a^2 = 4 => a = 2.

log_a(25) = log_a(5^2) = 2 log_a(5) = 2p, hence q = 2p.

log_a(25) = log_a(5^2) = 2 log_a(5) = 2p, hence q = 2p.

log_a(bc) = log_a(b) + log_a(c) = p + q.

log_a(bc) = log_a(b) + log_a(c) = p + q.

log_b(27) = 3 implies b^3 = 27 => b = 3.

log_x(16) = 4 implies x^4 = 16 => x^4 = 2^4 => x = 2.

log_x(27) = 3 implies x^3 = 27 => x = 3.

log_x(4) = 2 implies x^2 = 4 => x = 2.

log_x(81) = 4 implies x^4 = 81 => x^4 = 3^4 => x = 3.

log_x(81) = 4 implies x^4 = 81 => x = 3.

M · N = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

M · N = 2*3 + 2*3 + 2*3 = 6 + 6 + 6 = 18.

Substituting x = 2 into the equation gives 2^2 - 3*2 + p = 0 => 4 - 6 + p = 0 => p = 2.

Using the root, we substitute: 2^2 - 6*2 + k = 0 => 4 - 12 + k = 0 => k = 8.

Using Vieta's formulas, if one root is 3, the other root is 7 - 3 = 4. Thus, k = 3 * 4 = 12.

A relation is a function if every element in the domain (set A) maps to exactly one element in the codomain. Here, 1 maps to 2, and 2 maps to 3, but 3 has no mapping, so R is not a function.

A relation R is reflexive if every element in set A is related to itself. Since R contains (1, 1), (2, 2), and (3, 3), R is reflexive.

A relation is symmetric if for every (a, b) in R, (b, a) is also in R. Since R only contains pairs of the form (a, a), it is symmetric.

A relation R is transitive if whenever (a, b) ∈ R and (b, c) ∈ R, then (a, c) must also be in R. Here, (1, 2) and (2, 3) are in R, but (1, 3) is not, so R is not transitive.

R is not reflexive as (1,1), (2,2), (3,3) are not in R. It is symmetric as (2,3) implies (3,2) is not in R. It is transitive as (1,2) and (2,3) implies (1,3) is not in R. Thus, R is not all of the above.

R is reflexive because it contains all pairs (a, a) and symmetric because (1,2) implies (2,1).

R is not a partial order because it is not transitive; (1,2) and (2,2) do not imply (1,2).

R is reflexive and symmetric, but not transitive. Thus, both 1 and 2 are true.