\[(k^{x^2 + xy})(k^{y^2 + xy}) = k^{25}\]
In the equation above, k > 1 and x = 3. What is the positive value of y?
A) 1 B) 2 C) 4 D) 5
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).
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