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

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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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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]

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

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

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

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

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

n = 1
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Re: If k ≠ 0, k ≠ ±1, and (k^3*k*k^4)^2/k^n*k=k^14 , then n = [#permalink]
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 5315 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]
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