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# If f(x) = x^x,then f(f(x)) =

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Director
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If f(x) = x^x,then f(f(x)) =  [#permalink]

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23 Jun 2014, 01:51
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55% (hard)

Question Stats:

45% (00:45) correct 55% (00:31) wrong based on 397 sessions

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If $$f(x)= x^x,$$ then $$f(f(x))=$$

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Got this Question in Veritas prep test . Can this question be done by plugging in some values..

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If f(x) = x^x,then f(f(x)) =  [#permalink]

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23 Jun 2014, 02:25
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$$f(x) = x^x$$

$$f[f(x)] = f(x)^{f(x)}$$

$$= (x^x)^{(x^x)}$$

$$= x^{(x*x^x)}$$

$$= x^{x^{1+x}}$$

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Director
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Re: If f(x) = x^x,then f(f(x)) =  [#permalink]

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23 Jun 2014, 03:37
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PareshGmat wrote:
$$f(x) = x^x$$

$$f[f(x)] = f(x)^{f(x)}$$

$$= (x^x)^{(x^x)}$$

$$= x^{(x. x^x)}$$

$$= x^{x^{1+x}}$$

Would you try this question by plugging in. The question looked good for plugging in options but I am not getting the answer.
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If f(x) = x^x,then f(f(x)) =  [#permalink]

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Updated on: 25 Sep 2014, 18:40
2
1
WoundedTiger wrote:
PareshGmat wrote:
$$f(x) = x^x$$

$$f[f(x)] = f(x)^{f(x)}$$

$$= (x^x)^{(x^x)}$$

$$= x^{(x. x^x)}$$

$$= x^{x^{1+x}}$$

Would you try this question by plugging in. The question looked good for plugging in options but I am not getting the answer.

$$f(3) = 3^{3^{(1+3)}}$$

$$= 3^{(3^4)}$$

$$= 3^{81}$$........... (1)

$$f[f(3)] = (3^3)^{(3^3)}$$

$$= (3^3)^{27}$$

$$= 3^{(27*3)}$$

$$= 3^{81}$$ ............ (2)

(1) = (2)
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Originally posted by PareshGmat on 23 Jun 2014, 03:55.
Last edited by PareshGmat on 25 Sep 2014, 18:40, edited 1 time in total.
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If f(x) = x^x,then f(f(x)) =  [#permalink]

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23 Jun 2014, 04:07
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1

$$2^{3^2} = 2^{(3^2)} = 2^9 = 512$$

$$(2^3)^2 = 2^{(3*2)} = 2^6 = 64$$

OR

$$(2^3)^2 = 8^2 = 64$$

Power calculations go from top to bottom, if brackets are not provided
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Re: If f(x) = x^x,then f(f(x)) =  [#permalink]

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23 Jun 2014, 05:49
WoundedTiger wrote:
If $$f(x)= x^x,$$ then $$f(f(x))=$$

Attachment:
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Got this Question in Veritas prep test . Can this question be done by plugging in some values..

Added this question to functions directory in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions
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Re: If f(x) = x^x,then f(f(x)) =  [#permalink]

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07 Oct 2017, 08:07
Bunuel wrote:
WoundedTiger wrote:
If $$f(x)= x^x,$$ then $$f(f(x))=$$

Attachment:
Untitled.png

Got this Question in Veritas prep test . Can this question be done by plugging in some values..

Added this question to functions directory in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

hi Bunuel

the use of exponents is little bit trickier to me here

f [ f(x) ] to the power f(x)
that is [ f (x) ] is raised to power f(x)..

NOT, f (x) is raised to power f(x), if this is so, then it may mean, in f(x), "x" is raised to the power f(x), and thus may end up with result looking like option A

So, the expression here is meant to be "[ f (x) ]" (the whole) is raised to the power f(x)" and thus option D is right

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Re: If f(x) = x^x,then f(f(x)) =  [#permalink]

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07 Oct 2017, 08:20
gmatcracker2017 wrote:
Bunuel wrote:
WoundedTiger wrote:
If $$f(x)= x^x,$$ then $$f(f(x))=$$

Attachment:
Untitled.png

Got this Question in Veritas prep test . Can this question be done by plugging in some values..

Added this question to functions directory in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

hi Bunuel

the use of exponents is little bit trickier to me here

f [ f(x) ] to the power f(x)
that is [ f (x) ] is raised to power f(x)..

NOT, f (x) is raised to power f(x), if this is so, then it may mean, in f(x), "x" is raised to the power f(x), and thus may end up with result looking like option A

So, the expression here is meant to be "[ f (x) ]" (the whole) is raised to the power f(x)" and thus option D is right

Paresh HERE explains it so perfectly that I have nothing to add.
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Re: If f(x) = x^x,then f(f(x)) = &nbs [#permalink] 07 Oct 2017, 08:20
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