If a > 0, b = 2a, and c = 2b, which of the following is equivalent to a^b b^c?

1. Translate the problem requirements: We need to find what \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\) equals when \(\mathrm{a} > 0\), \(\mathrm{b} = 2\mathrm{a}\), and \(\mathrm{c} = 2\mathrm{b}\). The expression uses variables as both bases and exponents, so we need to substitute the given relationships systematically.
2. Express all variables in terms of a: Replace b and c with expressions involving only a to create a uniform base for comparison with answer choices.
3. Substitute into the target expression: Replace b and c in \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\) with their equivalent expressions in terms of a.
4. Apply exponent rules to simplify: Use properties of exponents to rewrite the expression in a form that matches one of the given answer choices.
5. Factor to match answer choice format: Reorganize the simplified expression to match the structure \((\text{base})^\mathrm{a}\) shown in all answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we're being asked to find. We have an expression \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\), and we need to figure out what this equals when we know that \(\mathrm{a} > 0\), \(\mathrm{b} = 2\mathrm{a}\), and \(\mathrm{c} = 2\mathrm{b}\).

Think of this like a chain of relationships: if a is some positive number, then b is twice that number, and c is twice b (which makes c four times a). We need to work out what happens when we raise a to the power of b, multiply that by b raised to the power of c, and see which answer choice this matches.

Process Skill: TRANSLATE - Converting the algebraic relationships into a clear understanding of what we need to calculate

2. Express all variables in terms of a

To make our calculations easier, let's write everything using just the variable a.

We're told:

• \(\mathrm{b} = 2\mathrm{a}\)
• \(\mathrm{c} = 2\mathrm{b}\)

Since \(\mathrm{b} = 2\mathrm{a}\), we can substitute this into the expression for c:
\(\mathrm{c} = 2\mathrm{b} = 2(2\mathrm{a}) = 4\mathrm{a}\)

So now we have everything in terms of a:

• \(\mathrm{b} = 2\mathrm{a}\)
• \(\mathrm{c} = 4\mathrm{a}\)

3. Substitute into the target expression

Now let's substitute these relationships into our original expression \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\).

Starting with \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\), we replace:

• b with \(2\mathrm{a}\)
• c with \(4\mathrm{a}\)

This gives us: \(\mathrm{a}^{(2\mathrm{a})} \times (2\mathrm{a})^{(4\mathrm{a})}\)

So our expression becomes: \(\mathrm{a}^{(2\mathrm{a})} \times (2\mathrm{a})^{(4\mathrm{a})}\)

4. Apply exponent rules to simplify

Now we need to work with \((2\mathrm{a})^{(4\mathrm{a})}\). We can use the exponent rule that says \((\mathrm{xy})^\mathrm{z} = \mathrm{x}^\mathrm{z} \times \mathrm{y}^\mathrm{z}\).

So \((2\mathrm{a})^{(4\mathrm{a})} = 2^{(4\mathrm{a})} \times \mathrm{a}^{(4\mathrm{a})}\)

Our full expression becomes:
\(\mathrm{a}^{(2\mathrm{a})} \times 2^{(4\mathrm{a})} \times \mathrm{a}^{(4\mathrm{a})}\)

Using the rule that \(\mathrm{x}^\mathrm{m} \times \mathrm{x}^\mathrm{n} = \mathrm{x}^{(\mathrm{m}+\mathrm{n})}\), we can combine the terms with base a:
\(\mathrm{a}^{(2\mathrm{a})} \times \mathrm{a}^{(4\mathrm{a})} = \mathrm{a}^{(2\mathrm{a} + 4\mathrm{a})} = \mathrm{a}^{(6\mathrm{a})}\)

So our expression is now: \(2^{(4\mathrm{a})} \times \mathrm{a}^{(6\mathrm{a})}\)

5. Factor to match answer choice format

Looking at our answer choices, they all have the format \((\text{something})^\mathrm{a}\). We need to rewrite our expression \(2^{(4\mathrm{a})} \times \mathrm{a}^{(6\mathrm{a})}\) in this form.

We can factor out the exponent a:
\(2^{(4\mathrm{a})} \times \mathrm{a}^{(6\mathrm{a})} = (2^4)^\mathrm{a} \times (\mathrm{a}^6)^\mathrm{a}\)

Since \(2^4 = 16\):
\((2^4)^\mathrm{a} \times (\mathrm{a}^6)^\mathrm{a} = 16^\mathrm{a} \times (\mathrm{a}^6)^\mathrm{a}\)

Using the rule that \(\mathrm{x}^\mathrm{n} \times \mathrm{y}^\mathrm{n} = (\mathrm{xy})^\mathrm{n}\):
\(16^\mathrm{a} \times (\mathrm{a}^6)^\mathrm{a} = (16 \times \mathrm{a}^6)^\mathrm{a} = (16\mathrm{a}^6)^\mathrm{a}\)

4. Final Answer

Our simplified expression is \((16\mathrm{a}^6)^\mathrm{a}\), which matches answer choice D exactly.

To verify: We started with \(\mathrm{a}^\mathrm{b} \times \mathrm{b}^\mathrm{c}\), substituted \(\mathrm{b} = 2\mathrm{a}\) and \(\mathrm{c} = 4\mathrm{a}\) to get \(\mathrm{a}^{(2\mathrm{a})} \times (2\mathrm{a})^{(4\mathrm{a})}\), simplified using exponent rules to get \(2^{(4\mathrm{a})} \times \mathrm{a}^{(6\mathrm{a})}\), and factored this as \((16\mathrm{a}^6)^\mathrm{a}\).

Answer: D. \((16\mathrm{a}^6)^\mathrm{a}\)