log_5(125) = log_5(5^3) = 3. Since log_5(25) = 2, we have x = 2, thus log_5(125) = 3.

log_5(5^x) = x.

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

log_5(x) = 1/2 implies x = 5^(1/2) = sqrt(5).

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

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

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.

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 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.

The equation sin^(-1)(x) + cos^(-1)(x) = π/2 holds for all x in the domain of the functions, which is [-1, 1]. Therefore, x can be any value in this range.

Using the identity sin^(-1)(x) + cos^(-1)(x) = π/2, we can conclude that x can take any value in the range [-1, 1].

If sin^(-1)(x) = π/4, then x = sin(π/4) = √2/2.

tan^(-1)(x) = π/4 implies that x = tan(π/4) = 1.

Setting the determinant to zero gives x*4 - 2*3 = 0, thus x = 1.5.

Substituting x = 1 gives 2(1)^2 + 3(1) + k = 0 => 2 + 3 + k = 0 => k = -5.