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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 03:34
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\(x^n + x^n + x^n = x^{(n + 1)}\), where x cannot equal zero. Which of the following must be true? I. n = 2 II. x = 3 III. n^x < x^n (A) I only (B) II and III (C) I and III (D) II only (E) None of the above
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 03:46
E for me. A is negated because when x equals 1, n can take any value. B is negated because x can take values other than 3. It can take 1. C is negated by combining cases taking x equals 1 and 3 for different values of n. E it is. Sent from my ONE A2003 using GMAT Club Forum mobile app



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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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Updated on: 22 Jul 2017, 11:43
i think it is D x^n + x^n + x^n = x^(n + 1) 3x^n=x*x^n x^n(3x)=0 as x is not equal to 0 only solution is x=3 i) we can not comment on value of n ii) we have got x= 3 iii) n^x < x^n put x= 3 (only solution of x) we get n^3 < 3^n for n =3 condition will not hold true.
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Originally posted by GTExl on 22 Jul 2017, 04:11.
Last edited by GTExl on 22 Jul 2017, 11:43, edited 5 times in total.



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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 07:16
Its D.. As \(3x^n=x^n*x\) and x is not 0 so x=3 Hence D. Press Kudos if this post helped you..!!



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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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Updated on: 22 Jul 2017, 11:05
Given : \(x^n + x^n + x^n = x^{(n+1)}\) = \(3* x^n = x^{n+1}\) = 3 = \(\frac{x^{n+1}}{x^n}\) = \(3^1\) = \(x^1\) Equating both sides we get x = 3. When x= 3, the value of n could be anything. I. n = 2; n COULD be 2. And not MUST be 2. II. x = 3. As described above X MUST be 3. III. \(n^x < x^n\). From the stem we have x=3. Substituting we get \(n^3 < 3^n\) . By checking for n = +1,1 and 0 we see the inquality holds true. But, does not hold true when n=3. So, this COULD be true. So only II Must be true and the answer is D
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Originally posted by susheelh on 22 Jul 2017, 09:39.
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 10:59
Bunuel wrote: \(x^n + x^n + x^n = x^{(n + 1)}\), where x cannot equal zero. Which of the following must be true?
I. n = 2 II. x = 3 III. n^x < x^n
(A) I only (B) II and III (C) I and III (D) II only (E) None of the above The answer is D Solving the equation gives x= 3, i.e 3 (x^n) = (x^n)x, x =3 Now, For I, n does not have to be 2 for x=3  Must not be true For II, x=3  Must be trueFor III, take a case where n=3=x, this negates n^x < x^n  Must not be true Hence answer is D.



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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 10:59
susheelh wrote: Given : \(x^n + x^n + x^n = x^{(n+1)}\)
= \(3* x^n = x^{n+1}\)
= 3 = \(\frac{x^{n+1}}{x^n}\)
= \(3^1\) = \(x^1\)
Equating both sides we get x = 3. When x= 3, the value of n could be anything.
I. n = 2; n COULD be 2. And not MUST be 2. II. x = 3. As described above X MUST be 3. III. \(n^x < x^n\). From the stem we have x=3. Substituting we get \(n^3 < 3^n\) . By checking for n = +1,1 and 0 we see the inquality holds true. So (Though a bit doubtful) , this MUST be true.
So II and III Must be true and IMO the answer is B What if n=3. then the third statement will be \(3^3<3^3\) (as x=3) which is not possible and the question stem gives no constraint for "n" Hence B is doubtful



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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 11:03
Excellent!!! I stand corrected. Will edit the response as well. Thanks niks18! (How could I even overlook that! ) niks18 wrote: susheelh wrote: Given : \(x^n + x^n + x^n = x^{(n+1)}\)
= \(3* x^n = x^{n+1}\)
= 3 = \(\frac{x^{n+1}}{x^n}\)
= \(3^1\) = \(x^1\)
Equating both sides we get x = 3. When x= 3, the value of n could be anything.
I. n = 2; n COULD be 2. And not MUST be 2. II. x = 3. As described above X MUST be 3. III. \(n^x < x^n\). From the stem we have x=3. Substituting we get \(n^3 < 3^n\) . By checking for n = +1,1 and 0 we see the inquality holds true. So (Though a bit doubtful) , this MUST be true.
So II and III Must be true and IMO the answer is B What if n=3. then the third statement will be \(3^3<3^3\) (as x=3) which is not possible and the question stem gives no constraint for "n" Hence B is doubtful
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 11:09
GTExl wrote: i think it is D x^n + x^n + x^n = x^(n + 1)
3x^n=x*x^n
x^n(3x)=0
as x is not equal to 0 only solution is x=3
i) we can not comment on value of n
ii) we have got x= 3
iii) n^x < x^n put x= 3 (only solution of x) we get n^3 < 3^n
for small value of n this true but for large vaue of n this is not true as 3^n is exponential.
For Ex
when n=1 1^3 <3^1 => 1<3 true
when n=5
5^3 < 3^5 => 125 < 243 False
Kudo if you find this solution right Hi, your answer is correct, but your explanation for why III must not be true is fautly i.e 125 < 243 is not False. The false case would be when n=3=x.



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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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22 Jul 2017, 11:37
Yes you are right. Thanks for telling me rulingbear. i have edited the post as well.
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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25 Jul 2017, 03:32
I picked D because X^(n+1) is X(X^n) so 3X^n = X(X^n) hence X=3



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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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30 Jul 2018, 09:21
We can simplify as 3x^n = x^(n+1) which equals to 3 = x for sure. So we can rid of options A, D and E. Then we need just to check statement III. Let say that n = 3. 3^3 < 3^3. Not true. Eliminating option B the remaning option is D.



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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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01 Aug 2018, 16:37
Bunuel wrote: \(x^n + x^n + x^n = x^{(n + 1)}\), where x cannot equal zero. Which of the following must be true?
I. n = 2 II. x = 3 III. n^x < x^n
(A) I only (B) II and III (C) I and III (D) II only (E) None of the above Simplifying we have: 3(x^n) = x^n * x^1 3 = x So, Roman numeral II must be true. Since the only choices involving II are B and D, we just need to test whether Roman numeral III is true. We note that 3^n + 3^n + 3^n = 3^(n+1) is satisfied when n = 3 (which also proves Roman numeral I is not necessarily true). But then, substituting x = n = 3 in Roman numeral III, we see that 3^3 < 3^3 is not true. Thus, among the given statements, only II must be true. Answer: D
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
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