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

With M opposite N and O to the right of M, P must be to the left of N, completing the arrangement.

With M to the right of N, we can visualize the arrangement. O's position opposite M helps us determine that Q is to the left of N.

M is the father of O, making M the father of N's father.

M is the aunt of O because she is the sister of O's father, N.

M is the uncle of N, which means M is the brother of N's parent O, making M the brother-in-law of O.

From M ≥ N and N > O, we can conclude M > O.

15 days from Monday is Wednesday.

For l=2, the orbital type is d.

O is the father of Q, and since Q is the sister of R, O is also the father of R.

The variable resistor allows for adjustment to achieve balance in the bridge.

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.

Using the root 2 in the equation: 2² - 6*2 + k = 0, we find k = 10.

Using the root, substitute x = 3: 3² - 7*3 + k = 0, which gives k = 10.

Using the root, substitute x = 3: 3² - 7*3 + k = 0, which gives k = 9.

Using the root, substitute x = 3: 3² - 7*3 + p = 0, which gives p = 6.

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.

Since Q = R and P > Q, it follows that P > R.