A ∩ B = {2} as 2 is the only even prime number.

det(A) = (1)(4) - (2)(3) = 4 - 6 = -2.

det(A) = (1*4) - (2*3) = 4 - 6 = -2.

det(2A) = 2^2 * det(A) = 4 * (1*5 - 2*3) = 4 * (-1) = -4.

Since A^2 = I, A is invertible because the inverse of A is A itself.

The determinant of a matrix is the product of its eigenvalues. Thus, det(A) = 1 * (-1) = -1.

The trace of a matrix is the sum of its eigenvalues. Therefore, trace(A) = 3 + 5 = 8.

The trace of a matrix is the sum of its eigenvalues. Thus, trace(A) = 2 + 3 + 4 = 9.

The equation can be formed as (x - 2)(x + 3) = 0, which expands to x^2 + x - 6 = 0.

det(B) = (2*4) - (3*1) = 8 - 3 = 5.

From cos^(-1)(x) = θ, we have cos(θ) = x. Therefore, sin(θ) = √(1 - cos^2(θ)) = √(1 - x^2).

log_10(20) = log_10(2 * 10) = log_10(2) + log_10(10) = a + 1.

log_10(x) = 2 implies x = 10^2 = 100.

log_2((x + 1)/x) = 1 implies (x + 1)/x = 2 => x + 1 = 2x => x = 1.

log_2(x + 1) = 3 implies x + 1 = 2^3 = 8 => x = 7.

log_2(x) + 2 = 5 => log_2(x) = 3 => x = 2^3 = 8.

log_2(x(x - 3)) = 3 => x(x - 3) = 2^3 = 8 => x^2 - 3x - 8 = 0. Solving gives x = 6.

log_2(x(x-1)) = 3 => x(x-1) = 2^3 = 8 => x^2 - x - 8 = 0. Solving gives x = 5.

log_2(x) = 5 implies x = 2^5 = 32.

log_3(9) = 2 and log_3(27) = 3, thus x = 2 + 3 = 5.

log_3(9) = log_3(3^2) = 2.

log_3(x + 1) = 2 implies x + 1 = 3^2 = 9 => x = 8.

log_3(4x) = 2 => 4x = 3^2 = 9 => x = 9/4 = 2.25.

log_3(x) = 2 implies x = 3^2 = 9.

log_4(64) = log_4(4^3) = 3.

log_4(x) = 1/2 implies x = 4^(1/2) = 2.

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

log_4(x) = 3 implies x = 4^3 = 64.

log_4(x) = 3 implies x = 4^3 = 64.

log_5(25) = 2 and log_5(5) = 1. Therefore, x = 2 + 1 = 3.