log_2(8) = 3 and log_2(4) = 2, thus x = 3 + 2 = 5.

log_2(x) + 2 = 6 implies log_2(x) = 4, so x = 2^4 = 16.

log_3(81) = log_3(3^4) = 4.

log_3(x) = 4 implies x = 3^4 = 81.

log_4(16) = log_4(4^2) = 2.

log_4(256) = log_4(4^4) = 4.

log_a(1) = 0 is true for any base a > 0, except when a = 1.

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

log_a(3) = 0.5 implies a^0.5 = 3, thus a = 3^2 = 9.

log_a(64) = 6 implies a^6 = 64. Since 64 = 2^6, we have a = 2.

log_b(25) = 2 implies b^2 = 25. Therefore, b = 5.

log_x(64) = 6 implies x^6 = 64. Since 64 = 2^6, we have x = 2.

The sum of the roots is 3 + (-1) = 2, hence p = -2.

Substituting x = 2 into the equation gives 2^2 - 4*2 + k = 0, which simplifies to k = 4.

Using the root 3 in the equation: 3^2 - 7*3 + k = 0, we get k = 6.

The double angle formula for cosine is cos 2A = cos²A - sin²A.

Using the double angle formula, sin 2A = 2(1/2)(√(1 - (1/2)²)) = 2(1/2)(√(3/4)) = 1.

sin 2(30°) = sin 60° = √3/2.

Sin A = 0 at A = 0°, 180°, and 360°. The principal value is 0°.

Using the Pythagorean identity, cos A = √(1 - sin²A) = √(1 - 0.36) = √(0.64) = 0.8.

Using the identity tan A = sin A / cos A. First, find cos A using cos A = √(1 - sin² A) = √(1 - 0.64) = 0.6. Therefore, tan A = 0.8 / 0.6 = 1.3333, which is approximately 1.2.

The angle A for which sin A = 1/2 is π/6 radians.

In the first quadrant, sin A = 1/2 corresponds to A = 30°.

tan A = sin A / cos A = (1/√2) / (1/√2) = 1.

Using the Pythagorean identity, cos A = √(1 - sin²A) = √(1 - (5/13)²) = √(1 - 25/169) = √(144/169) = 12/13.

Using the Pythagorean identity, cos C = √(1 - sin² C) = √(1 - 0.25) = √0.75 = √3/2.

Using the Pythagorean identity, cos C = √(1 - sin²C) = √(1 - 0.8²) = √(1 - 0.64) = √(0.36) = 0.6.

Using the Pythagorean identity, cos x = √(1 - sin²x) = √(1 - 0.6²) = √(1 - 0.36) = √(0.64) = 0.8.

Using the Pythagorean identity, cos θ = √(1 - sin²θ) = √(1 - 0.8²) = √(1 - 0.64) = √(0.36) = 0.6.

Using the identity, sin²A = 1 - cos²A = 1 - (0.6)² = 1 - 0.36 = 0.64.