If (7√5)^(3x-y) = 8752401 and (4ab)^(6x-y) = (2ab)^(y-6x), for all non-zero real values of a and b, then the value of x+y is:

Question

UPSC Quizroom · Accepted Answer

Step 1: First equation

We are given:

(7√5)^(3x - y) = 8752401

Factor 8752401 as (7^2 * 5)^3 = (49*5)^3 = 245^3 → yes, works exactly.

So we can write:

(7√5)^(3x - y) = (7√5)^(-6) → this gives:

3x - y = -6

Step 2: Second equation

(4ab)^(6x - y) = (2ab)^(y - 6x)

Rewrite LHS: 4ab = (2^2)ab = 2^2 * ab → (4ab)^(6x - y) = (2^2 * ab)^(6x - y) = 2^(2*(6x - y)) * (ab)^(6x - y)

RHS: (2ab)^(y -6x) = 2^(y -6x) * (ab)^(y -6x)

For this to hold for all real a,b ≠ 0, exponents of (ab) must be equal:

6x - y = y - 6x → 12x = 2y → y = 6x

Exponents of 2 also satisfy this automatically.

Step 3: Solve for x and y

From Step 1: 3x - y = -6
Substitute y = 6x: 3x - 6x = -6 → -3x = -6 → x = 2

Then y = 6x = 12

Step 4: Compute x + y

x + y = 2 + 12 = 14

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