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The value of (8^4 + 8^(16))/(4^8 + 16^8) is

GMAT Algebra : (ALG) Questions

Source: Mock
Algebra
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The value of \(\frac{8^4 + 8^{16}}{4^8 + 16^8}\) is

A
less than \(0.00005\)
B
greater than \(0.00005\) and less than \(0.05\)
C
greater than \(0.05\) and less than \(50\)
D
greater than \(50\) and less than \(50{,}000\)
E
greater than \(50{,}000\)
Solution
  1. Translate the problem requirements: We need to find the numerical value of the fraction \(\frac{8^4 + 8^{16}}{4^8 + 16^8}\) and determine which range it falls into among the given answer choices.
  2. Express all terms using a common base: Convert all numbers (8, 4, 16) to powers of 2 to simplify the expression and make calculations manageable.
  3. Factor and simplify the expression: Look for common factors in the numerator and denominator that can be canceled out to reduce the complexity.
  4. Estimate the final value: Calculate or approximate the simplified expression to determine which answer choice range it falls into, noting that the ranges are quite wide so precise calculation may not be necessary.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to find. We have a fraction with two parts:

  • The top (numerator): \(8^4 + 8^{16}\)
  • The bottom (denominator): \(4^8 + 16^8\)

We need to calculate this fraction's value and see which of the five given ranges it falls into. Notice that the ranges are quite spread out - from very small decimals (less than 0.00005) to very large numbers (greater than 50,000). This suggests we might not need an exact calculation, just a good approximation.

Process Skill: TRANSLATE

2. Express all terms using a common base

Here's the key insight: all the numbers in our expression (8, 4, and 16) can be written as powers of 2. This will make our calculations much easier.

Let's convert each number:

  • \(8 = 2^3\) (since \(2 \times 2 \times 2 = 8\))
  • \(4 = 2^2\) (since \(2 \times 2 = 4\))
  • \(16 = 2^4\) (since \(2 \times 2 \times 2 \times 2 = 16\))

Now let's rewrite our original expression using these:

  • \(8^4 = (2^3)^4 = 2^{12}\)
  • \(8^{16} = (2^3)^{16} = 2^{48}\)
  • \(4^8 = (2^2)^8 = 2^{16}\)
  • \(16^8 = (2^4)^8 = 2^{32}\)

So our fraction becomes: \(\frac{2^{12} + 2^{48}}{2^{16} + 2^{32}}\)

Process Skill: SIMPLIFY

3. Factor and simplify the expression

Now we can look for common factors to cancel out. Let's factor out the smallest power of 2 from both the numerator and denominator.

In the numerator \((2^{12} + 2^{48})\), the smaller power is \(2^{12}\), so:

\(2^{12} + 2^{48} = 2^{12}(1 + 2^{36}) = 2^{12}(1 + 2^{36})\)

In the denominator \((2^{16} + 2^{32})\), the smaller power is \(2^{16}\), so:

\(2^{16} + 2^{32} = 2^{16}(1 + 2^{16}) = 2^{16}(1 + 2^{16})\)

Our fraction becomes:

\(\frac{2^{12}(1 + 2^{36})}{2^{16}(1 + 2^{16})} = \frac{(1 + 2^{36})}{2^4(1 + 2^{16})}\)

Since \(2^4 = 16\), we have:

\(\frac{(1 + 2^{36})}{16(1 + 2^{16})}\)

4. Estimate the final value

Now let's think about the relative sizes of these numbers:

  • \(2^{16} = 65,536\) (a large number)
  • \(2^{36}\) is astronomically larger than \(2^{16}\)

Since \(2^{36}\) is much, much larger than 1, we can approximate:

\(1 + 2^{36} \approx 2^{36}\)

Since \(2^{16}\) is much larger than 1, we can approximate:

\(1 + 2^{16} \approx 2^{16}\)

So our fraction is approximately:

\(\frac{2^{36}}{16 \times 2^{16}} = \frac{2^{36}}{2^4 \times 2^{16}} = \frac{2^{36}}{2^{20}} = 2^{16}\)

Since \(2^{16} = 65,536\), our answer is approximately 65,536.

Process Skill: INFER

5. Final Answer

Our calculated value of approximately 65,536 falls in the range "greater than 50,000", which corresponds to answer choice E.

To double-check: 65,536 > 50,000, so it definitely fits in the highest range provided.

Answer: E

Common Faltering Points

Errors while devising the approach

  • Attempting direct calculation: Students may try to calculate the exact values of large numbers like \(8^{16}\) and \(16^8\) directly, leading to unwieldy computations. They miss the key insight that converting to a common base (powers of 2) will dramatically simplify the problem.
  • Overlooking the range-based answer choices: Students may not notice that the answer choices are given as ranges rather than exact values. This means they can use approximation techniques rather than needing precise calculations, which would save significant time and effort.
  • Missing the common base connection: Students might not recognize that 4, 8, and 16 are all powers of 2. Without this insight, they cannot effectively simplify the expression and may get stuck with complex arithmetic.

Errors while executing the approach

  • Exponent multiplication errors: When converting expressions like \(8^4 = (2^3)^4\), students frequently make mistakes with the exponent rules, incorrectly calculating \((2^3)^4\) as \(2^7\) (adding exponents) instead of \(2^{12}\) (multiplying exponents).
  • Incorrect factoring: When factoring out common terms from expressions like \(2^{12} + 2^{48}\), students may factor incorrectly or choose the wrong power to factor out, leading to errors in simplification.
  • Approximation misjudgment: Students might incorrectly approximate terms, such as keeping the '+1' terms when they should recognize that \(2^{36} \gg 1\) and \(2^{16} \gg 1\), making these '+1' terms negligible for estimation purposes.

Errors while selecting the answer

  • Calculation verification miss: Students may correctly calculate \(2^{16} = 65,536\) but fail to properly verify which range this falls into, potentially confusing the ranges or misreading the inequalities in the answer choices.
Answer Choices Explained
A
less than \(0.00005\)
B
greater than \(0.00005\) and less than \(0.05\)
C
greater than \(0.05\) and less than \(50\)
D
greater than \(50\) and less than \(50{,}000\)
E
greater than \(50{,}000\)
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