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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.
Last edited by susheelh on 22 Jul 2017, 11:05, edited 1 time in total.
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 true For III, take a case where n=3=x, this negates n^x < x^n - Must not be true
Re: x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the
[#permalink]
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22 Jul 2017, 10:59
1
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
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
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
1
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.
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.
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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