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The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is

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The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is  [#permalink]

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New post 31 May 2018, 21:28
1
\({\frac{1}{5}*(2^{-14}+2^{-15}+2^{-16}+2^{-17})}\) = K * \(2^{-17}\)

To simplify multiply both sides by \(2^{17}\) to get:-

\({\frac{1}{5}*(2^3+2^2+2^1+1)}\) = K * 1[/m]

This implies k = 3 (Correct Answer)
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Re: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is  [#permalink]

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New post 29 Aug 2019, 05:43
Bunuel wrote:
andih wrote:
The value of (2^-14)+(2^-15)+(2^-16) + (2^-17) is how times the value of 2^-17?

A. 3/2

B. 5/2

C. 3

D. 4

E. 5


Original question reads:
The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is how many times the value of 2^(-17)?

We need to find the value of: \(\frac{\frac{1}{5}*(2^{-14}+2^{-15}+2^{-16}+2^{-17})}{ 2^{-17}}=\frac{\frac{1}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})}{\frac{1}{2^{17}}}\).

Now, \(\frac{\frac{1}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})}{\frac{1}{2^{17}}}=\frac{2^{17}}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})=\frac{1}{5}*(2^3+2^2+2+1)=\frac{1}{5}*15=3\).

Answer: C.

I followed up until the 3rd to the last step. I am confused on how you got \(\frac{1}{5}\)


Can someone please help me out, how did you go from \(\frac{2^{17}}{5}*(\frac{1}{2^{14}}+\frac{1}{2^{15}}+\frac{1}{2^{16}}+\frac{1}{2^{17}})=\frac{1}{5}*(2^3+2^2+2+1)\)
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Re: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is  [#permalink]

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New post 03 Mar 2020, 11:12
The value of (2(−14)+2(−15)+2(−16)+2(−17))/5 is how many times the value of 2(−17)2(−17)?

A. 3/2
B. 5/2
C. 3
D. 4
E. 5

First off, we can factor out a 2^-17 from the top of the fraction, which leaves us (2^-17)(2^3+2^2+2^1+1) in the numerator. The whole fraction then turns into ((2^-17)*15)/5, which turns into 3(2^-17), which is 3 times 2^-17. Hope this helps.
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Re: The value of (2^(-14) + 2^(-15) + 2^(-16) + 2^(-17))/5 is   [#permalink] 03 Mar 2020, 11:12

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