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If 1/5^n = 1/5^k + 1/5^k + 1/5^k + 1/5^k + 1/5^k, then n expressed in

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If 1/5^n = 1/5^k + 1/5^k + 1/5^k + 1/5^k + 1/5^k, then n expressed in  [#permalink]

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New post 25 Mar 2018, 23:26
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Question Stats:

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If \(\frac{1}{5^n} = \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k}\), then n expressed in terms of k is

A. k - 1
B. k + 1
C. 1 - k
D. 3k
E. k^5

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If 1/5^n = 1/5^k + 1/5^k + 1/5^k + 1/5^k + 1/5^k, then n expressed in  [#permalink]

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New post 26 Mar 2018, 00:36
Bunuel wrote:
If \(\frac{1}{5^n} = \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k}\), then n expressed in terms of k is

A. k - 1
B. k + 1
C. 1 - k
D. 3k
E. k^5


\(\frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} = \frac{5}{5^k}\)

\(\frac{1}{5^n} = \frac{5^1}{5^k}\) -> \(5^{-n} = 5^{1-k}\) -> \(-n = 1-k\) -> \(n = k-1\) (multiplying by -1 on both sides)

Therefore, n expressed in terms of k is k-1(Option A)
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Re: If 1/5^n = 1/5^k + 1/5^k + 1/5^k + 1/5^k + 1/5^k, then n expressed in  [#permalink]

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New post 27 Mar 2018, 10:36
Bunuel wrote:
If \(\frac{1}{5^n} = \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k} + \frac{1}{5^k}\), then n expressed in terms of k is

A. k - 1
B. k + 1
C. 1 - k
D. 3k
E. k^5


Simplifying the equation, we have:

1/5^n = 5/5^k

Cross-multiplying, we obtain:

5^k = 5^n x 5

5^k = 5^(n + 1)

Since the bases are equal, we can equate the exponents:

k = n + 1

k - 1 = n

Answer: A
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Re: If 1/5^n = 1/5^k + 1/5^k + 1/5^k + 1/5^k + 1/5^k, then n expressed in &nbs [#permalink] 27 Mar 2018, 10:36
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