\[(k^{x^2 + xy})(k^{y^2 + xy}) = k^{25}\]

In the equation above, k > 1 and x = 3. What is the positive value of y?

A) 1

B) 2

C) 4

D) 5

A) 1

B) 2

C) 4

D) 5

When multiplying variables with the same base and different exponents, add the exponents. The equation above becomes:

\[k^{x^2 + xy + y^2 + xy} = k^{25}\]

Combine like terms to get:

\[k^{x^2 + 2xy + y^2} = k^{25}\]

Therefore,

\[x^2 + 2xy + y^2 = 25\]

If you factor the quadratic you get

\[(x + y)^2 = 25\]

\[x + y = \pm 5\]

Substitute for x = 3 to get 3 + y = +/- 5. Since the question asks for positive value of y, we have 3 + y = 5, or y = 2.

**The correct answer is (B).**