For three positive integers p, q, r, x^(pq²) = y^(qr) = z^(p²r) and r = pq + 1 such that 3, 3 log_yx, 3 log_zy, 7 logₓz are in A.P. with common difference 1/2. Then r-p-q is equal to

Solution

Maths sir · 5 steps

Set the three powers equal to one constant k, and each log ratio becomes a simple ratio of p, q, r.

1. Read the AP. First term 3 and common difference 1/2 give the terms 3, 7/2, 4, 9/2. So 3 log_y x = 7/2, 3 log_z y = 4 and 7 logₓ z = 9/2. That gives log_y x = 7/6, log_z y = 4/3 and logₓ z = 9/14.
2. Take logs. Take natural logs and call the common value k: pq² ln x = qr ln y = p² r ln z = k. Divide one log by another, and k cancels in each ratio.

   log_y x = qr/pq² = r/pq, log_z y = (p² r)/qr = p²/q, logₓ z = pq²/(p² r) = q²/pr

3. Solve the system. From r/pq = 7/6 with r = pq + 1, we get (pq+1)/pq = 7/6, so pq = 6 and r = 7. From p²/q = 4/3 with q = 6/p, we get 3p² = 24/p, so p³ = 8. That gives p = 2 and q = 3.
4. Check. The third ratio is q²/pr = 9/14, which matches the fourth AP term. All three values fit.
5. So. r - p - q = 7 - 2 - 3 = 2, which is option (d).

The idea to keep. When several powers are equal, set each one equal to k and take logs. Every log of one base to another then becomes a plain ratio of the exponents.

Answer from the NTA final answer key. Solution by PrepWiser.