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16 Sep 2014, 00:11
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What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} }\)?
A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)
A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)
16 Sep 2014, 00:11
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Official Solution:
What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} }\)?
A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)
Simplify the given expression by expressing all values as powers of 2:
\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} } = \)
\(=\frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4} } = \)
\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)
\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)
\(= \frac{1}{\frac{1}{2^4} } = \)
\(= 2^4=\)
\(= 16\)
Answer: A
What is the value of \(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} }\)?
A. 16
B. 4
C. 2
D. \(\frac{1}{2}\)
E. \(\frac{1}{16}\)
Simplify the given expression by expressing all values as powers of 2:
\(\frac{\frac{1}{8} * (\frac{1}{16})^2 * 4^4}{(\frac{1}{64})^{\frac{1}{2} } * 2^{-4} } = \)
\(=\frac{\frac{1}{2^3} * \frac{1}{2^8} * 2^8}{\frac{1}{8} * \frac{1}{2^4} } = \)
\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)
\(= \frac{\frac{1}{2^3} }{\frac{1}{2^3} * \frac{1}{2^4} } = \)
\(= \frac{1}{\frac{1}{2^4} } = \)
\(= 2^4=\)
\(= 16\)
Answer: A
Arro44
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08 Jun 2018, 03:56
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Quote:
Thank you for elaborating.
I somehow have issues with such problems on a frequent basis.
Does breaking the respective fraction down into the respective prime and exponent represent the best go to solution?
