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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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Bunuel
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x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink] New post   Updated on: 09 Dec 2019, 18:08
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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, 01:30.
Last edited by chetan2u on 09 Dec 2019, 18:08, edited 1 time in total.
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x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink] New post   09 Dec 2019, 09:19
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\(x^n + x^n + x^n = x^{n+1}........3*x^n=x^n*x.........x^n(3-x)=0\)
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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nick1816
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Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink] New post   09 Dec 2019, 11:06
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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 \(x^n+x^n+x^n=x^{n+1}\)
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Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink] New post   16 Apr 2024, 00:21
chetan2u wrote:
\(x^n + x^n + x^n = x^{n+1}........3*x^n=x^n*x.........x^n(3-x)=0\)
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 \(x^n+x^n+x^n=x^{n+1}\)

­why is st 2 insufficient. i believe n would1 acccording it it
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Bunuel
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Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink] New post   16 Apr 2024, 03:06
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Its_me_aka_ak wrote:
chetan2u wrote:
\(x^n + x^n + x^n = x^{n+1}........3*x^n=x^n*x.........x^n(3-x)=0\)
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 \(x^n+x^n+x^n=x^{n+1}\)

­why is st 2 insufficient. i believe n would1 acccording it it

­
If x = 3, then we get:

\(3^n + 3^n + 3^n = 3^{n+1}\)

\(3*3^n = 3^{n+1}\)

\(3^{n+1} = 3^{n+1}\)

The above is true for any value of n. Hence, (2) is not sufficient.

Hope it's clear.
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Re: x^n + x^n + x^n + x^n = x^(n + 1). What is the value of n? [#permalink]
16 Apr 2024, 03:06
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