The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?
A. 3/2
B. 5/2
C. 3
D. 4
E. 5
C
OG Negative Exponents Q
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Say the value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is x times the value of 2^(−17).AbeNeedsAnswers wrote:The value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?
A. 3/2
B. 5/2
C. 3
D. 4
E. 5
C
We then need to get the value of x.
=> [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)] / 5 = 2^(−17)*x
Taking 2^(−17) common from the RHS. We get,
2^(−17)[[2^(3) + 2^(2) + 2^(1) + 1] / 5] = 2^(−17)*x
=> [2^(3) + 2^(2) + 2^(1) + 1] / 5 = x; 2^(−17) gets cancelled
= [8 + 4 + 2 + 1] /5 = x
= [15] /5 = x
[spoiler]x = 3[/spoiler].
The correct answer: C
Hope this helps!
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Let's rephrase the question first. We're asked to solve for x in the following:
(2�¹� + 2�¹� + 2�¹� + 2�¹�)/5 = x * 2�¹�
Let's multiply both sides by 5:
(2�¹� + 2�¹� + 2�¹� + 2�¹�) = 5x * 2�¹�
Then divide both sides by 2�¹�:
2³ + 2² + 2 + 1 = 5x
which simplifies to
3 = x
(2�¹� + 2�¹� + 2�¹� + 2�¹�)/5 = x * 2�¹�
Let's multiply both sides by 5:
(2�¹� + 2�¹� + 2�¹� + 2�¹�) = 5x * 2�¹�
Then divide both sides by 2�¹�:
2³ + 2² + 2 + 1 = 5x
which simplifies to
3 = x
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We can create the following equation in which k is an integer:AbeNeedsAnswers wrote: ↑Wed Jul 19, 2017 10:04 pmThe value of [2^(−14) + 2^(−15) + 2^(−16) + 2^(−17)]/ 5 is how many times the value of 2^(−17)?
A. 3/2
B. 5/2
C. 3
D. 4
E. 5
C
[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] / 5 = (k)(2^-17)
[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] = (k)(2^-17)(5)
Let’s multiply each side of the equation by 2^17:
2^17[2^(-14) + 2^(-15) + 2^(-16) + 2^(-17)] = (2^17)(2^-17)(k)(5)
2^3 + 2^2 + 2^1 + 1 = 5k
(8 + 4 + 2 + 1) = 5k
15 = 5k
3 = k
Answer: C
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