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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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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the  [#permalink]

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New post 22 Jul 2017, 03:34
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A
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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  [#permalink]

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New post 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.

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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the  [#permalink]

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New post 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(3-x)=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  [#permalink]

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New post 22 Jul 2017, 07:16
3
Its D..
As \(3x^n=x^n*x\) and x is not 0
so x=3
Hence D.

:-D

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x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the  [#permalink]

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New post 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.
Last edited by susheelh on 22 Jul 2017, 11:05, edited 1 time in total.
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Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the  [#permalink]

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New post 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 true
For 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  [#permalink]

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New post 22 Jul 2017, 10:59
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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  [#permalink]

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New post 22 Jul 2017, 11:03
Excellent!!! I stand corrected. Will edit the response as well. Thanks niks18! (How could I even overlook that! :shock: )

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  [#permalink]

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New post 22 Jul 2017, 11:09
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GTExl wrote:
i think it is D
x^n + x^n + x^n = x^(n + 1)

3x^n=x*x^n

x^n(3-x)=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  [#permalink]

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New post 22 Jul 2017, 11:37
:shock: 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  [#permalink]

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New post 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  [#permalink]

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New post 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  [#permalink]

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New post 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 &nbs [#permalink] 01 Aug 2018, 16:37
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