If n is a positive integer, then \(\frac{(8^{2\mathrm{n}})(3^{-4\mathrm{n}})}{(2^{-\mathrm{n}})(9^{-5\mathrm{n}})} =\)

1. Translate the problem requirements: We need to simplify the expression \(\frac{(8^{2n})(3^{-4n})}{(2^{-n})(9^{-5n})}\) by converting all terms to prime base factors and applying exponent rules
2. Convert composite bases to prime factors: Express 8 and 9 in terms of their prime factors (2 and 3 respectively) to work with consistent bases
3. Apply exponent rules for division: Use the rule that \(\frac{a^m}{a^n} = a^{m-n}\) to combine like bases in numerator and denominator
4. Simplify and match answer format: Combine the final exponents and express in the form that matches the given answer choices

Execution of Strategic Approach

1. Translate the problem requirements

We need to simplify the expression \(\frac{(8^{2n})(3^{-4n})}{(2^{-n})(9^{-5n})}\). Let's think about what this means in plain English: we have a fraction where the top part (numerator) contains 8 raised to the power 2n multiplied by 3 raised to the power -4n, and the bottom part (denominator) contains 2 raised to the power -n multiplied by 9 raised to the power -5n.

The key insight is that we want to work with the same base numbers throughout our calculation. Since 8 and 9 are not prime numbers, we should convert them to their prime factor forms so we can use the rules for combining exponents.

Process Skill: TRANSLATE - Converting the complex expression into manageable components with a clear plan

2. Convert composite bases to prime factors

Let's break down the composite numbers into their prime factors. Think of this like taking apart building blocks to see what they're really made of:

• 8 = 2 × 2 × 2 = \(2^3\)
• 9 = 3 × 3 = \(3^2\)

Now let's rewrite each term:

• \(8^{2n} = (2^3)^{2n} = 2^{6n}\) (using the power rule: when we raise a power to a power, we multiply the exponents)
• \(9^{-5n} = (3^2)^{-5n} = 3^{-10n}\)

So our expression becomes:
\(\frac{(2^{6n})(3^{-4n})}{(2^{-n})(3^{-10n})}\)

3. Apply exponent rules for division

Now we can group terms with the same base together. When we divide numbers with the same base, we subtract the exponents. Think of this like canceling out common factors:

For the base 2 terms:

• In the numerator: \(2^{6n}\)
• In the denominator: \(2^{-n}\)
• Combined: \(2^{6n} ÷ 2^{-n} = 2^{6n - (-n)} = 2^{6n + n} = 2^{7n}\)

For the base 3 terms:

• In the numerator: \(3^{-4n}\)
• In the denominator: \(3^{-10n}\)
• Combined: \(3^{-4n} ÷ 3^{-10n} = 3^{-4n - (-10n)} = 3^{-4n + 10n} = 3^{6n}\)

Process Skill: SIMPLIFY - Systematically combining like terms using exponent rules

4. Simplify and match answer format

Putting our results together, we get:
\(2^{7n} × 3^{6n} = (2^{7n})(3^{6n})\)

Let's verify this matches one of our answer choices:

1. \((2^{5n})(3^{-14n})\) - No, our exponents are different
2. \((2^{7n})(3^{6n})\) - Yes! This exactly matches our result
3. \(\frac{2^{4n}}{3^{n}}\) - No, different exponents and format
4. \(\frac{2^{5n}}{3^{6n}}\) - No, different exponents and format
5. \(\frac{4^{3n}}{3^{n}}\) - No, different base and exponents

Final Answer

The answer is B: \((2^{7n})(3^{6n})\)

We successfully simplified the original complex expression by converting composite bases to prime factors and applying exponent rules systematically. Our final result matches answer choice B exactly.

Common Faltering Points

Errors while devising the approach

1. Not recognizing the need to convert composite bases to prime factors
Many students attempt to work directly with bases like 8 and 9 without converting them to \(2^3\) and \(3^2\). This makes it nearly impossible to apply exponent rules correctly since you can't combine terms with different bases. Students might get stuck trying to simplify \(8^{2n} ÷ 2^{-n}\) without realizing they need a common base.

2. Misunderstanding how to handle negative exponents in fractions
Students often get confused about whether negative exponents in the denominator should be treated as positive or negative when moving terms around. For example, seeing \(2^{-n}\) in the denominator, they might incorrectly think this becomes \(-2^n\) instead of understanding that dividing by \(2^{-n}\) is the same as multiplying by \(2^n\).

Errors while executing the approach

1. Arithmetic errors when applying the power rule
When converting \(8^{2n}\) to \((2^3)^{2n}\), students frequently make mistakes multiplying the exponents, getting \(2^{5n}\) or \(2^{8n}\) instead of the correct \(2^{6n}\). Similarly, with \(9^{-5n} = (3^2)^{-5n}\), they might get \(3^{-7n}\) or \(3^{-3n}\) instead of \(3^{-10n}\).

2. Sign errors when subtracting negative exponents
In the division step, students often struggle with \(6n - (-n)\) or \(-4n - (-10n)\). They might forget to change subtraction of a negative to addition, getting results like \(2^{5n}\) instead of \(2^{7n}\), or \(3^{-14n}\) instead of \(3^{6n}\). This is especially tricky because both the exponent and the sign can be negative.

3. Incorrectly applying the quotient rule for exponents
Students sometimes add exponents instead of subtracting when dividing, or they might subtract in the wrong order. For example, when computing \(2^{6n} ÷ 2^{-n}\), they might calculate \(2^{6n + (-n)} = 2^{5n}\) instead of \(2^{6n - (-n)} = 2^{7n}\).

Errors while selecting the answer

1. Choosing an answer that matches only one part of their result
Students who correctly get \(2^{7n}\) and \(3^{6n}\) might panic when they see their work and hastily select answer choice A because it has powers of both 2 and 3, without carefully checking that the exponents match. They might focus on the format (both bases present) rather than the specific values of the exponents.