If 2^(x) + 2^(x) + 2^(x) + 2^(x) = 2^(n), what is the value of x in terms of n?

1. Translate the problem requirements: We have four identical terms (\(2^x\)) being added together, and this sum equals \(2^n\). We need to find what x equals in terms of n.
2. Simplify the left side using addition: Recognize that adding four identical terms is the same as multiplying by 4, so \(2^x + 2^x + 2^x + 2^x = 4 \times 2^x\).
3. Express 4 as a power of 2: Convert 4 to base 2 format (\(4 = 2^2\)) so both sides of the equation have the same base.
4. Apply exponent rules to combine terms: Use the rule that \(a^m \times a^n = a^{(m+n)}\) to simplify \(4 \times 2^x = 2^2 \times 2^x = 2^{(2+x)}\).
5. Set exponents equal and solve: Since \(2^{(2+x)} = 2^n\), the exponents must be equal, giving us \(2+x = n\), so \(x = n-2\).

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we have here. We're given an equation where four identical terms are added together: \(2^x + 2^x + 2^x + 2^x\), and this sum equals \(2^n\). Our job is to figure out what x equals when we express it in terms of n.

Think of it this way - imagine you have four boxes, each containing the same amount (which is \(2^x\)), and when you add up everything in all four boxes, you get a total of \(2^n\). We need to find out what's in each individual box (the value of x) based on the total (which involves n).

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical relationship

2. Simplify the left side using addition

Now, when we add four identical things together, that's the same as multiplying that thing by 4. It's like saying "I have 4 groups of the same item" instead of "I have this item plus this item plus this item plus this item."

So: \(2^x + 2^x + 2^x + 2^x = 4 \times 2^x\)

This is much cleaner to work with. Now our equation becomes: \(4 \times 2^x = 2^n\)

3. Express 4 as a power of 2

Here's the key insight: since we're working with powers of 2, we want everything in our equation to have the same base (base 2). Right now we have the number 4 on the left side, but 4 can be written as a power of 2.

Since \(2 \times 2 = 4\), we know that \(4 = 2^2\).

So our equation becomes: \(2^2 \times 2^x = 2^n\)

Now both sides of our equation are expressed in terms of powers of 2, which makes our work much easier.

4. Apply exponent rules to combine terms

When we multiply powers that have the same base, we add the exponents. This is a fundamental rule: \(a^m \times a^n = a^{(m+n)}\).

So: \(2^2 \times 2^x = 2^{(2+x)}\)

Our equation now looks like: \(2^{(2+x)} = 2^n\)

This is much simpler - we have the same base (2) on both sides, just with different exponents.

5. Set exponents equal and solve

Here's the beautiful part: when we have the same base on both sides of an equation, the exponents must be equal for the equation to be true.

Since \(2^{(2+x)} = 2^n\), we must have: \(2+x = n\)

Solving for x: \(x = n - 2\)

Final Answer

Therefore, \(x = n - 2\)

Looking at our answer choices, this matches choice D: \(n - 2\).

We can verify this makes sense: if \(x = n - 2\), then each term \(2^x = 2^{(n-2)}\), and four of them gives us \(4 \times 2^{(n-2)} = 2^2 \times 2^{(n-2)} = 2^n\) ✓

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the equation structure

Students might not recognize that \(2^x + 2^x + 2^x + 2^x\) represents four identical terms being added together. They may try to work with each term separately or attempt to combine them incorrectly, missing the key insight that this simplifies to \(4 \times 2^x\).

2. Not recognizing the need for a common base

Students may attempt to solve the equation without converting everything to powers of 2. They might try to work with \(4 \times 2^x = 2^n\) directly without realizing that expressing 4 as \(2^2\) is crucial for applying exponent rules effectively.

Errors while executing the approach

1. Incorrect application of exponent rules

When combining \(2^2 \times 2^x\), students might incorrectly multiply the exponents (getting \(2^{(2\times x)} = 2^{(2x)}\)) instead of adding them. The correct rule is \(a^m \times a^n = a^{(m+n)}\), so \(2^2 \times 2^x = 2^{(2+x)}\).

2. Algebraic errors when solving for x

After correctly establishing that \(2 + x = n\), students might make simple algebraic mistakes such as getting \(x = n + 2\) instead of \(x = n - 2\), or forgetting to isolate x properly.

3. Incorrect simplification of the left side

Students might incorrectly combine the four terms, perhaps writing \(2^x + 2^x + 2^x + 2^x = 2^{(4x)}\) by mistakenly applying exponent rules to addition, rather than recognizing this as \(4 \times 2^x\).

Errors while selecting the answer

No likely faltering points - once students have correctly solved for \(x = n - 2\), the answer choice D is clearly identifiable and matches exactly.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a convenient value for n
Let's use \(n = 4\), which gives us \(2^n = 2^4 = 16\). This choice makes our calculations manageable since 16 is easily divisible by 4.

Step 2: Set up the equation with our chosen value
Substituting \(n = 4\) into our original equation:
\(2^x + 2^x + 2^x + 2^x = 16\)

Step 3: Simplify the left side
Since we have four identical terms being added:
\(4 \times 2^x = 16\)

Step 4: Solve for \(2^x\)
Dividing both sides by 4:
\(2^x = 16 \div 4 = 4\)

Step 5: Find the value of x
Since \(2^x = 4\) and \(4 = 2^2\), we have:
\(x = 2\)

Step 6: Test answer choices
With \(n = 4\) and \(x = 2\), let's check each option:

1. \(n/4 = 4/4 = 1 \neq 2\) ❌
2. \(4n = 4(4) = 16 \neq 2\) ❌
3. \(2n = 2(4) = 8 \neq 2\) ❌
4. \(n - 2 = 4 - 2 = 2\) ✓
5. \(n + 1 = 4 + 1 = 5 \neq 2\) ❌

Step 7: Verify with another value
Let's confirm with \(n = 5\):
\(4 \times 2^x = 2^5 = 32\)
\(2^x = 8 = 2^3\), so \(x = 3\)
Checking option D: \(n - 2 = 5 - 2 = 3\) ✓

Conclusion: The answer is D. \(n - 2\)