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by sparkle6 » Wed Sep 21, 2011 7:59 am
If f is the function defined for all k such that f(k) = k^5/16, what is f(2k) in terms of f(k)?

a. 1/8 f(k)
b. 5/8 f(k)
c. 2 f(k)
d. 10 f(k)
e. 32 f(k)
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by cans » Wed Sep 21, 2011 8:14 am
f(k) = k^5/16, what is f(2k) in terms of f(k)??
f(2k) = (2k)^5/16 = 2^5 * k^5/16 = 32*f(k)
IMO E
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by sl750 » Wed Sep 21, 2011 12:01 pm
f(k)=k^5/16.
f(2k)=(2k)^5/16
Shouldn't it be (2^5/16)*(k^5/16)

(32)^1/16*f(k)
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by gmatclubmember » Wed Sep 21, 2011 12:14 pm
sl750 wrote:f(k)=k^5/16.
f(2k)=(2k)^5/16
Shouldn't it be (2^5/16)*(k^5/16)

(32)^1/16*f(k)
You are right... it should be 32/16 * f(k) = 2f(k).
OA is C

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by sl750 » Wed Sep 21, 2011 12:16 pm
More like 16th root of 32 *f(k)
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by gmatclubmember » Wed Sep 21, 2011 12:23 pm
sl750 wrote:More like 16th root of 32 *f(k)
No I think the question is k raised to the power 5 and then this whole value divided by 16.
and NOT k raised to the power 5/16.

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by sparkle6 » Sun Sep 25, 2011 6:21 am
The correct answer is E
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