x^n + x^n + x^n = x^(n + 1), where x cannot equal zero. Which of the : Problem Solving (PS)

GTExl

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Updated on: 22 Jul 2017, 11:43

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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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.

susheelh

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Updated on: 22 Jul 2017, 11:05

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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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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22 Jul 2017, 10:59

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Bunuel

$x^n + x^n + x^n = x^{(n + 1)}$, 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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22 Jul 2017, 10:59

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susheelh

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

susheelh

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22 Jul 2017, 11:03

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Excellent!!! I stand corrected. Will edit the response as well. Thanks niks18! (How could I even overlook that! :shock:

)

niks18

susheelh

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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25 Jul 2017, 03:32

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I picked D because X^(n+1) is X(X^n) so 3X^n = X(X^n) hence X=3

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30 Jul 2018, 09:21

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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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01 Aug 2018, 16:37

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Bunuel

$x^n + x^n + x^n = x^{(n + 1)}$, 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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10 Aug 2020, 13:10

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I solved this by simplifying the question stem:

$x^{n} + x^{n} + x^{n} = x^{n+1} $
$x^{n}(1 + 1 + 1) = x^{n+1}$
$x^{n}(3) = x^{n}(x^{1})$
$x = 3$

Since x = 3 was answer II, B and D were only items to choose from. Since I could not determine on any information regarding n, I eliminated B and went with D.

Bunuel

$x^n + x^n + x^n = x^{(n + 1)}$, 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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19 Jun 2025, 09:43

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Hi why can't we take x as 1 and if we can why does2 have to be true

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19 Jun 2025, 14:08

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kayarat600

Hi why can't we take x as 1 and if we can why does2 have to be true

$x^n + x^n + x^n = x^{(n + 1)}$

Simplifying the above,

=> $x^n * (1 + 1 + 1) = x^n * x$

=> $x^n * 3 = x^n * x$

From above, you get x = 3

If you were to take x = 1, then LHS becomes 3, and RHS becomes 1 => and LHS is not equal to RHS, hence it would be invalid to take x = 1.

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

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