h(1) = 1^3 - 3*1 = 1 - 3 = -2.

I · J = 1*2 + 1*2 + 1*2 = 2 + 2 + 2 = 6.

I · J = 1*3 + 2*2 + 3*1 = 3 + 4 + 3 = 10.

I · J = 1*4 + 2*5 + 3*6 = 4 + 10 + 18 = 32.

I · J = 2a + 2b + 2c = 12 gives the equation 2a + 2b + 2c = 12.

If 'a' is excluded, we can form subsets from {b, c}, which has 2^2 = 4 subsets.

log(2x) = 3 => 2x = 10^3 => x = 1000.

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.

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).