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What is the value of x?
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03 Jan 2019, 20:34
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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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____________________________ Regards, Chaitanya +1 Kudos if you like my explanation!!!



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Re: What is the value of x?
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04 Jan 2019, 09:31
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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Re: What is the value of x?
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04 Jan 2019, 12:42
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 multiindexes, 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?
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04 Jan 2019, 20:01
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 multiindexes, 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.
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Re: What is the value of x?
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04 Jan 2019, 22:54
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 multiindexes, 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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Re: What is the value of x?
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05 Jan 2019, 17:11
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 pointofview".) Thank you all for your interest in the theme, Regards, Fabio.
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Re: What is the value of x?
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