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# x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n?

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Math Expert
Joined: 02 Sep 2009
Posts: 61403
x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n?  [#permalink]

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Updated on: 09 Dec 2019, 17:08
00:00

Difficulty:

45% (medium)

Question Stats:

54% (02:00) correct 46% (01:57) wrong based on 46 sessions

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

What is the value of n?

(1) $$x^n = 9\sqrt{3}$$

(2) $$x = 3$$

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Originally posted by Bunuel on 09 Dec 2019, 00:30.
Last edited by chetan2u on 09 Dec 2019, 17:08, edited 1 time in total.
Corrected the question
Math Expert
Joined: 02 Aug 2009
Posts: 8252
x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n?  [#permalink]

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09 Dec 2019, 08:19
$$x^n + x^n + x^n = x^{n+1}........3*x^n=x^n*x.........x^n(3-x)=0$$
Thus either x^n=0 or x=3

What is the value of n?

(1) $$x^n = 9\sqrt{3}$$
So x=3as $$x^n\neq{0}$$
So $$3^n=9\sqrt{3}$$..
We can find n from it.
$$3^n=9\sqrt{3}=3^{5/2}$$..n=5/2

(2) $$x = 3$$
Insufficient

A

The question should be $$x^n+x^n+x^n=x^{n+1}$$
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Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n?  [#permalink]

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09 Dec 2019, 10:06
You are absolutely correct.

At x=3, x^n(4-x)=0 can never be 0 for real values of n.

Question is flawed imo.

chetan2u wrote:
$$x^n + x^n + x^n + x^n = x^{n+1}........4*x^n=x^n*x.........x^n(4-x)=0$$
Thus either x^n=0 or x=4

What is the value of n?

(1) $$x^n = 9\sqrt{3}$$
So x=4 as $$x^n\neq{0}$$
So $$4^n=9\sqrt{3}$$..
We can find n from it.
However the question should have x^n three times only, such that $$3^n=9\sqrt{3}=3^{5/2}$$..n=5/2

(2) $$x = 3$$
Insufficient

A

The question should be $$x^n+x^n+x^n=x^{n+1}$$
Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n?   [#permalink] 09 Dec 2019, 10:06
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