If \(\frac{2^{(\mathrm{x}+\mathrm{y})^2}}{2^{(\mathrm{x}-\mathrm{y})^2}} = 2\), what is the value of xy?

1. Translate the problem requirements: We have a fraction with powers of 2 in both numerator and denominator that equals 2. We need to find the value of the product xy.
2. Apply exponent rules to simplify the fraction: Use the rule that dividing powers with the same base means subtracting exponents to reduce the complex fraction to a single power of 2.
3. Set up equation using the simplified form: Since we know the simplified expression equals 2, we can write it as a power of 2 and use the fact that if the bases are equal, the exponents must be equal.
4. Expand and solve for the target variable: Expand the squared terms, simplify the resulting expression, and solve directly for xy.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have here. We're given a fraction where both the top and bottom are powers of 2, and this whole fraction equals 2.

In everyday terms, think of this like having two different "strengths" of the number 2 - one divided by the other - and the result is just 2 itself. We need to find what xy equals.

The expression is: \(\frac{2^{(\mathrm{x}+\mathrm{y})^2}}{2^{(\mathrm{x}-\mathrm{y})^2}} = 2\)

Our goal is to find the value of the product xy.

Process Skill: TRANSLATE

2. Apply exponent rules to simplify the fraction

When we divide numbers with the same base, we subtract the exponents. Think of it this way: if you have \(2^5 ÷ 2^3\), you're essentially canceling out three 2's from both top and bottom, leaving you with \(2^2\).

So here, when we divide \(2^{(\mathrm{x}+\mathrm{y})^2}\) by \(2^{(\mathrm{x}-\mathrm{y})^2}\), we subtract the exponents:

\(\frac{2^{(\mathrm{x}+\mathrm{y})^2}}{2^{(\mathrm{x}-\mathrm{y})^2}} = 2^{(\mathrm{x}+\mathrm{y})^2 - (\mathrm{x}-\mathrm{y})^2}\)

Now our equation becomes:

\(2^{(\mathrm{x}+\mathrm{y})^2 - (\mathrm{x}-\mathrm{y})^2} = 2\)

Process Skill: SIMPLIFY

3. Set up equation using the simplified form

Since we know that \(2^1 = 2\), and we have \(2^{(\mathrm{x}+\mathrm{y})^2 - (\mathrm{x}-\mathrm{y})^2} = 2\), this means the exponents must be equal.

When the bases are the same (both are 2), the only way the equation can be true is if the exponents are equal:

\((\mathrm{x}+\mathrm{y})^2 - (\mathrm{x}-\mathrm{y})^2 = 1\)

Process Skill: INFER

4. Expand and solve for the target variable

Now we need to expand those squared terms. Let's work through this step by step:

\((\mathrm{x}+\mathrm{y})^2 = \mathrm{x}^2 + 2\mathrm{xy} + \mathrm{y}^2\)

\((\mathrm{x}-\mathrm{y})^2 = \mathrm{x}^2 - 2\mathrm{xy} + \mathrm{y}^2\)

Substituting back into our equation:

\((\mathrm{x}^2 + 2\mathrm{xy} + \mathrm{y}^2) - (\mathrm{x}^2 - 2\mathrm{xy} + \mathrm{y}^2) = 1\)

Let's distribute the negative sign:

\(\mathrm{x}^2 + 2\mathrm{xy} + \mathrm{y}^2 - \mathrm{x}^2 + 2\mathrm{xy} - \mathrm{y}^2 = 1\)

Notice how the \(\mathrm{x}^2\) terms cancel out, and the \(\mathrm{y}^2\) terms cancel out:

\(2\mathrm{xy} + 2\mathrm{xy} = 1\)

\(4\mathrm{xy} = 1\)

Therefore: \(\mathrm{xy} = \frac{1}{4}\)

Process Skill: MANIPULATE

4. Final Answer

The value of xy is \(\frac{1}{4}\), which corresponds to answer choice (D).

We can verify this makes sense: our algebraic manipulation led us from a complex exponential equation to a simple linear equation in xy, and the answer is one of the given choices.

Common Faltering Points

Errors while devising the approach

1. Misapplying exponent rules initially
Students might try to solve this by cross-multiplying or treating it as a regular algebraic fraction, forgetting that when bases are the same, we can use the rule \(\frac{\mathrm{a}^\mathrm{m}}{\mathrm{a}^\mathrm{n}} = \mathrm{a}^{\mathrm{m}-\mathrm{n}}\). This leads them down a much more complicated path.

2. Not recognizing the target variable
Some students may get distracted trying to solve for individual values of x and y separately, rather than recognizing that the question specifically asks for the product xy. This can lead to unnecessary work and potential errors.

Errors while executing the approach

1. Sign errors when expanding \((\mathrm{x}-\mathrm{y})^2\)
When expanding \((\mathrm{x}-\mathrm{y})^2\), students commonly write it as \(\mathrm{x}^2 - 2\mathrm{xy} + \mathrm{y}^2\) but then make sign errors when subtracting this entire expression. They might forget to distribute the negative sign properly, writing \(-(\mathrm{x}^2 - 2\mathrm{xy} + \mathrm{y}^2)\) as \(-\mathrm{x}^2 - 2\mathrm{xy} - \mathrm{y}^2\) instead of the correct \(-\mathrm{x}^2 + 2\mathrm{xy} - \mathrm{y}^2\).

2. Arithmetic errors in the final calculation
After correctly getting \(4\mathrm{xy} = 1\), students might make simple division errors, writing \(\mathrm{xy} = 4\) instead of \(\mathrm{xy} = \frac{1}{4}\), or incorrectly simplifying the fraction.

3. Forgetting that \(2^1 = 2\)
Students might correctly simplify to \(2^{(\mathrm{x}+\mathrm{y})^2 - (\mathrm{x}-\mathrm{y})^2} = 2\) but then struggle to recognize that this means the exponent must equal 1, since \(2^1 = 2\). They might try to take logarithms or use other unnecessary approaches.

Errors while selecting the answer

No likely faltering points - the final calculation directly gives xy = 1/4, which clearly corresponds to answer choice (D).