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# If f is a function defined for all k by f(k) = k^5/16, what is f(2k)

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Director
Joined: 27 Dec 2004
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If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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04 Jul 2005, 08:12
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Difficulty:

25% (medium)

Question Stats:

67% (01:10) correct 33% (01:40) wrong based on 441 sessions

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If f is a function defined for all k by $$f(k) = \frac{k^5}{16}$$, what is f(2k) in terms of f(k)?

A. $$\frac{1}{8} f(k)$$

B. $$\frac{5}{8} f(k)$$

C. $$2 f(k)$$

D. $$10 f(k)$$

E. $$32 f(k)$$
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Joined: 09 Jun 2005
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Location: Hong Kong
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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05 Jul 2005, 10:02
E for me.

F(k) = k^5 / 16

F(2k) = (2k)^5 / 16
= {(2^5) * (k^5)} / 16
= 32 (k^5) / 16
= 32 F(k)
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Posts: 1155
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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05 Jul 2005, 11:14
Folaa3 wrote:
If f is a function defined for all k by 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)

E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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05 Jul 2005, 16:55
banerjeea_98 wrote:
E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)

Baner, Can you explain the difference between C & E
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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06 Jul 2005, 08:14
banerjeea_98 wrote:
E is correct.

f(k) = k^5/16.......f(2k) = (2k)^5/16 ---> 32 k^5/16 ==> 32 f(k)

Baner, Can you explain the difference between C & E

When a function f(k) is defined as in this case as = k^5/16 , that means that the function is valid for all values of k. So when you are asked f(2k), what u do is simply replace K with 2K. In this case u do this
{(2k)^5}/16 ===> 32 k^/16 ==> 32 f(k)
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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24 Aug 2011, 06:12
f(k) = k^5/16
f(2k) = (2k)^5/16 = 32k^5/16 = 32 x (k^5/16) = 32f(k).
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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24 Aug 2011, 21:06
f(k) = k^5/16

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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16 May 2015, 03:37
Jasonammex wrote:
If f is the function defined for all k by 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)

I find this the easier with plugging in, instead of trying to solve with all the variables:

Let's assume k = 1, so f(k)=k^5/16 = f(1)=1/16.

The question now asks what is f(2k) in terms of f(k):

f(2k) = 2^5/16 = 32/16 (or 2, but not important to reduce)

Since f(k) = 1/16 and f(2k) = 32/16 f(2k) in terms of f(k) is 32f(k)
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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16 Mar 2018, 04:14
Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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16 Mar 2018, 06:20
teamryan15 wrote:
Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?

Hey teamryan15 ,

Looks like you messed up something there!

Let me explain you:

$$f(2k) = (2k)^5/16$$ = [$$(2)^5 * (k)^5$$]/16 = 2 * $$k^5$$ -- (1)

Now, you know that f(k) = $$k^5$$ / 16

=> $$k^5$$ = 16 f(k) -- (2)

Substitute the value at (2) to (1)

You will get, f(2k) = 32 f(k)

Does that make sense?
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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16 Mar 2018, 15:09
abhimahna wrote:
teamryan15 wrote:
Can someone explain this to me?

I understand the steps until the very end.

f(2k) = (2k)^5/16 = 2k^5 = 2 * 16*k^5 = 32 f(k)

Shouldnt it be : 2k^5 = 32*K^5 Instead of just 32 f(k)? Whats happening to the exponent for k?

Hey teamryan15 ,

Looks like you messed up something there!

Let me explain you:

$$f(2k) = (2k)^5/16$$ = [$$(2)^5 * (k)^5$$]/16 = 2 * $$k^5$$ -- (1)

Now, you know that f(k) = $$k^5$$ / 16

=> $$k^5$$ = 16 f(k) -- (2)

Substitute the value at (2) to (1)

You will get, f(2k) = 32 f(k)

Does that make sense?

Yes it does thanks.
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Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)  [#permalink]

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21 Mar 2018, 01:00
Re: If f is a function defined for all k by f(k) = k^5/16, what is f(2k)   [#permalink] 21 Mar 2018, 01:00
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