Question:medium

\(\frac{2×4×8×16}{(log_24)^2(log_48)^3(log_816)^4}\) equals [This Question was asked as TITA]

Updated On: Jan 15, 2026
  • 25
  • 24
  • 22
  • 23
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The Correct Option is B

Solution and Explanation

To simplify the expression \(\frac{2×4×8×16}{(log_24)^2(log_48)^3(log_816)^4}\), we will evaluate its components.

First, calculate the individual logarithms in the denominator:

\(log_24 = 2\) as \(2^2 = 4\).

\(log_48\). We know \(8 = 2^3\) and \(log_28 = 3\). Using the change of base formula, \(log_48 = \frac{log_28}{log_24} = \frac{3}{2}\).

\(log_816\). We know \(16 = 2^4\) and \(log_216 = 4\). Using the change of base formula, \(log_816 = \frac{log_216}{log_28} = \frac{4}{3}\).

Substitute these values back into the original expression:

\(\frac{2×4×8×16}{(log_24)^2(log_48)^3(log_816)^4} = \frac{2×4×8×16}{(2)^2(\frac{3}{2})^3(\frac{4}{3})^4}\)

Calculate each part:

Numerator: \(2 × 4 × 8 × 16 = 1024\).

Denominator components: \(2^2 = 4\), \((\frac{3}{2})^3 = \frac{27}{8}\), and \((\frac{4}{3})^4 = \frac{256}{81}\).

Combine the denominator components:

Denominator: \(4 × \frac{27}{8} × \frac{256}{81}\).

Simplify the fraction:

\(\frac{1024}{4 × \frac{27}{8} × \frac{256}{81}} = \frac{1024}{\frac{4 × 27 × 256}{8 × 81}} = \frac{1024}{\frac{27648}{648}}\).

Reduce the fraction in the denominator: \(\frac{27648}{648} = 42.666...\). There appears to be a calculation error in the provided steps. Revisiting the calculation of \(\frac{log_28}{log_24}\) yields \(\frac{3}{2}\). For \(\frac{log_216}{log_28}\) it yields \(\frac{4}{3}\). The calculation of the denominator should be \(4 \times (\frac{3}{2})^3 \times (\frac{4}{3})^4 = 4 \times \frac{27}{8} \times \frac{256}{81}\). Simplifying this gives \( \frac{4 \times 27 \times 256}{8 \times 81} = \frac{1 \times 1 \times 256}{2 \times 3} = \frac{256}{6} = \frac{128}{3}\). Thus, the expression is \(\frac{1024}{\frac{128}{3}} = 1024 \times \frac{3}{128} = 8 \times 3 = 24\).

Therefore, the expression simplifies to:

\(\frac{1024}{24} = \boxed{24}\).

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