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Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  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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Math Expert V
Joined: 02 Sep 2009
Posts: 58402
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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Difficulty:   25% (medium)

Question Stats: 83% (00:50) correct 17% (01:40) wrong based on 60 sessions

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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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Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3335
Location: India
GPA: 3.12
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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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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Target Test Prep Representative D
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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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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

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If you find one of my posts helpful, please take a moment to click on the "Kudos" button. 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] 27 Mar 2018, 11:36
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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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