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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =

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Math Expert V
Joined: 02 Sep 2009
Posts: 58381
If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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Difficulty:   5% (low)

Question Stats: 86% (01:18) correct 14% (01:45) wrong based on 131 sessions

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If k ≠ 0, k ≠ ±1, and $$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$ , then n =

A. -1
B. 1
C. 3
D. 49
E. 129

Kudos for a correct solution.

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Director  Joined: 07 Aug 2011
Posts: 502
Concentration: International Business, Technology
GMAT 1: 630 Q49 V27 If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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Bunuel wrote:
If k ≠ 0, k ≠ ±1, and $$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$ , then n =

A. -1
B. 1
C. 3
D. 49
E. 129

Kudos for a correct solution.

$$K^{16-n-1} = K^{14}$$

n=1
Retired Moderator B
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 300
Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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Bunuel wrote:
If k ≠ 0, k ≠ ±1, and $$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$ , then n =

A. -1
B. 1
C. 3
D. 49
E. 129

Kudos for a correct solution.

The powers will be (3+1+4)2-n-1=14
15-n=14
n=1

Intern  Joined: 22 Aug 2014
Posts: 40
Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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1
Bunuel wrote:
If k ≠ 0, k ≠ ±1, and $$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$ , then n =

A. -1
B. 1
C. 3
D. 49
E. 129

Kudos for a correct solution.

Deducing the eqn, we get:

(K^8)^2/K^(n+1) = K*14

16 - (n + 1) = 14
n = 1

Option B
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If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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Answer = B = 1

$$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$

$$\frac{(k^8)^2}{k^n * k^1}=k^{14}$$

$$\frac{k^{16}}{k^n * k^1}=k^{14}$$

$$\frac{k^{15}}{k^n}=k^{14}$$

$$k^n = k^{1}$$

n = 1
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Joined: 02 Sep 2009
Posts: 58381
Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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Bunuel wrote:
If k ≠ 0, k ≠ ±1, and $$\frac{(k^3*k*k^4)^2}{k^n*k}=k^{14}$$ , then n =

A. -1
B. 1
C. 3
D. 49
E. 129

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:
Attachment: determinetheexponentII_text.PNG [ 14.37 KiB | Viewed 2288 times ]

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Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =  [#permalink]

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_________________ Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n =   [#permalink] 03 Nov 2018, 02:03
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