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What is the value of x?

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What is the value of x?  [#permalink]

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New post 03 Jan 2019, 19:34
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What is the value of x?

1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0

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Chaitanya


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Re: What is the value of x?  [#permalink]

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New post 04 Jan 2019, 08:31
eswarchethu135 wrote:
What is the value of x?

1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0


1) \(x^x = 1\)
Let x be 0, \(0^0=1\), also 1^1=0
Thus x acb be 0 or 1
Insufficient

2) \(x^a = 1\) where a > 0
\(1^a=1\) and \((-1)^a\), when a is even..
Thus x can be 1 or -1..
Insufficient

Combined
x is 1..
Sufficient

C
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: What is the value of x?  [#permalink]

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New post 04 Jan 2019, 11:42
eswarchethu135 wrote:
What is the value of x?

1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0

chetan2u ´s argument for the insufficiency of statement (2) is perfect.

The SUFFICIENCY of statement (1) is proved below, although I must say the arguments are OUT of GMAT´s scope (in terms of abstraction, so to speak).

Important: 0^0 is not equal to 1, it´s an undefined mathematical expression. (If you prefer, it does not exist, from the fact that it cannot be defined properly.)

The reason people believe it is equal to 1 is related to the fact that in SOME cases it is convenient to use 0^0=1 as a CONVENTION.

(The best example I know is when dealing with multi-indexes, in the context of Partial Differential Equations, for instance.)

All that put, let´s see the reason why (1) is SUFFICIENT and, therefore, the correct answer is (A):


\({x^x} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{x^x}} \right| = 1\,\,\,\,\,\left( * \right)\)

\(x > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,0 < x < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,0 < x < 1\,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr
\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{possible}} \hfill \cr
\,x > 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} > \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,x > 1\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

\(x < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = {1 \over {{x^{\left| x \right|}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\mathop = \limits^{\left( * \right)} \,\,\left| {{x^x}} \right|\, = {1 \over {\,\left| {{x^{\left| x \right|}}} \right|}} = {1 \over {\left| x \right|{\,^{\left| x \right|}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| x \right|{\,^{\left| x \right|}}\,\,\, = \,\,\,1\,\,\,\,\left\{ \matrix{
\,\,\,\left| x \right| \ne 1\,\,\,\,\mathop \Rightarrow \limits^{{\rm{previous}}\,\,{\rm{cases}}} \,\,\,\,\,{\rm{impossible}} \hfill \cr
\,\,\,x = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^x} = - 1 \ne 1 \hfill \cr} \right.\)


Regards,
Fabio.
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Re: What is the value of x?  [#permalink]

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New post 04 Jan 2019, 19:01
1
fskilnik wrote:
eswarchethu135 wrote:
What is the value of x?

1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0

chetan2u ´s argument for the insufficiency of statement (2) is perfect.

The SUFFICIENCY of statement (1) is proved below, although I must say the arguments are OUT of GMAT´s scope (in terms of abstraction, so to speak).

Important: 0^0 is not equal to 1, it´s an undefined mathematical expression. (If you prefer, it does not exist, from the fact that it cannot be defined properly.)

The reason people believe it is equal to 1 is related to the fact that in SOME cases it is convenient to use 0^0=1 as a CONVENTION.

(The best example I know is when dealing with multi-indexes, in the context of Partial Differential Equations, for instance.)

All that put, let´s see the reason why (1) is SUFFICIENT and, therefore, the correct answer is (A):


\({x^x} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{x^x}} \right| = 1\,\,\,\,\,\left( * \right)\)

\(x > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,0 < x < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,0 < x < 1\,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr
\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{possible}} \hfill \cr
\,x > 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} > \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,x > 1\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

\(x < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = {1 \over {{x^{\left| x \right|}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\mathop = \limits^{\left( * \right)} \,\,\left| {{x^x}} \right|\, = {1 \over {\,\left| {{x^{\left| x \right|}}} \right|}} = {1 \over {\left| x \right|{\,^{\left| x \right|}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| x \right|{\,^{\left| x \right|}}\,\,\, = \,\,\,1\,\,\,\,\left\{ \matrix{
\,\,\,\left| x \right| \ne 1\,\,\,\,\mathop \Rightarrow \limits^{{\rm{previous}}\,\,{\rm{cases}}} \,\,\,\,\,{\rm{impossible}} \hfill \cr
\,\,\,x = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^x} = - 1 \ne 1 \hfill \cr} \right.\)


Regards,
Fabio.


I would argue against someone claiming 0^0 is not 1.
Yes, 0^0 can be called undefined at some places but 0^0 is taken as 1 in many important theorems and maths would break down if we did not accept this fact..
For example..
Binomial expansion..
\(7^2=(7+0)^2=7^20^0+7^10^1+7^00^2=49+0+0=49\)

If we take 0^0 as undefined \(7^20^0\) should be undefined and not 49..

However 0^0 being debatable between 1 and undefined will not be tested on GMAT.
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: What is the value of x?  [#permalink]

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New post 04 Jan 2019, 21:54
chetan2u wrote:
fskilnik wrote:
eswarchethu135 wrote:
What is the value of x?

1) \(x^x = 1\)
2) \(x^a = 1\) where a > 0

chetan2u ´s argument for the insufficiency of statement (2) is perfect.

The SUFFICIENCY of statement (1) is proved below, although I must say the arguments are OUT of GMAT´s scope (in terms of abstraction, so to speak).

Important: 0^0 is not equal to 1, it´s an undefined mathematical expression. (If you prefer, it does not exist, from the fact that it cannot be defined properly.)

The reason people believe it is equal to 1 is related to the fact that in SOME cases it is convenient to use 0^0=1 as a CONVENTION.

(The best example I know is when dealing with multi-indexes, in the context of Partial Differential Equations, for instance.)

All that put, let´s see the reason why (1) is SUFFICIENT and, therefore, the correct answer is (A):


\({x^x} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\left| {{x^x}} \right| = 1\,\,\,\,\,\left( * \right)\)

\(x > 0\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\{ \matrix{
\,0 < x < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} < 1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,0 < x < 1\,\,\,\,\,{\rm{impossible}}\,\, \hfill \cr
\,x = 1\,\,\,\,\, \Rightarrow \,\,\,\,{\rm{possible}} \hfill \cr
\,x > 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} > \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = 1\,\,,\,\,\,x > 1\,\,\,\,\,{\rm{impossible}} \hfill \cr} \right.\)

\(x < 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^x} = {1 \over {{x^{\left| x \right|}}}}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\mathop = \limits^{\left( * \right)} \,\,\left| {{x^x}} \right|\, = {1 \over {\,\left| {{x^{\left| x \right|}}} \right|}} = {1 \over {\left| x \right|{\,^{\left| x \right|}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left| x \right|{\,^{\left| x \right|}}\,\,\, = \,\,\,1\,\,\,\,\left\{ \matrix{
\,\,\,\left| x \right| \ne 1\,\,\,\,\mathop \Rightarrow \limits^{{\rm{previous}}\,\,{\rm{cases}}} \,\,\,\,\,{\rm{impossible}} \hfill \cr
\,\,\,x = - 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{x^x} = - 1 \ne 1 \hfill \cr} \right.\)


Regards,
Fabio.


I would argue against someone claiming 0^0 is not 1.
Yes, 0^0 can be called undefined at some places but 0^0 is taken as 1 in many important theorems and maths would break down if we did not accept this fact..
For example..
Binomial expansion..
\(7^2=(7+0)^2=7^20^0+7^10^1+7^00^2=49+0+0=49\)

If we take 0^0 as undefined \(7^20^0\) should be undefined and not 49..

However 0^0 being debatable between 1 and undefined will not be tested on GMAT.


There is a debate still going on about \(0^0\) is equal to 1 or undefined. But in GMAT sense and also in daily math usage anything power 0 is 1. This applies to \(0^0\) as well. So to avoid confusion its better to accept \(0^0\) as 1, atleast in GMAT.
_________________

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Regards,

Chaitanya


+1 Kudos

if you like my explanation!!!

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Re: What is the value of x?  [#permalink]

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New post 05 Jan 2019, 16:11
In the GMAT the test taker will not be in the position to have to choose between the two options (1 or undefined).

I am aware that in many cases it is useful to consider 0^0 = 1, as I explained in my previous post. Cheetan´s example is a good one, by the way.

On the other hand, one cannot define 0^0 as 1 (or as any other real number) without coming into math contradictions, specially in Mathematical Analysis (the area in which I am a Ph.D. candidate). In other words, accepting 1 is the possibility that "gets into trouble", not the other choice.

(That´s why considering 0^0 undefined is the "conservative point-of-view".)

Thank you all for your interest in the theme,

Regards,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net

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Re: What is the value of x? &nbs [#permalink] 05 Jan 2019, 16:11
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