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What is the least integer p for which 27^p > 3^18?

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What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 21 Mar 2018, 23:07
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A
B
C
D
E

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Re: What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 21 Mar 2018, 23:11
Bunuel wrote:
What is the least integer p for which \(27^p > 3^{18}\)?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 18


3^3p > 3^18
p>6
p=7
B
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What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 21 Mar 2018, 23:15
Bunuel wrote:
What is the least integer p for which \(27^p > 3^{18}\)?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 18



\(27^p\) can be simplified as \((3^3)^p = 3^{3p}\) because\((a^m)^n = a^{m*n}\)

\(3^{3p} > 3^{18}\) -> \(3p > 18\) -> \(p > 6\)

Therefore, the least value of p for which \(27^p > 3^{18}\) is 7 (Option B)
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What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 22 Mar 2018, 01:41

Solution



We need to find the smallest integer p for which \(27^p >3^{18}\).

Working out:

    • \(27^p >3^{18}\)

    • \((3^3)^p > 3^{18}\)

    • By applying \({(a^m)}^n = a^{m*n}\), we get \(3^{3p} > 3^{18}\)

    • Hence, \(3p> 18\).
      o \(p>6.\)

    • Hence, the smallest integer value of \(p\) greater than \(6\) is 7.

Answer: B
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Re: What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 23 Mar 2018, 10:16
Bunuel wrote:
What is the least integer p for which \(27^p > 3^{18}\)?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 18


\(27^p > 3^{18}\)

\(3^3p > 3^{18}\)

3p > 18.....p>6

Least integer power is 7

Answer: B
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Re: What is the least integer p for which 27^p > 3^18?  [#permalink]

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New post 23 Mar 2018, 12:41
Bunuel wrote:
What is the least integer p for which \(27^p > 3^{18}\)?

(A) 6
(B) 7
(C) 8
(D) 9
(E) 18


We first re-express 27 as 3^3, and so 27^p = (3^3)^p = 3^3p. Simplifying the inequality, we have:

27^p > 3^18

3^3p > 3^18

3p > 18

p > 6

The least integer greater than 6 is 7.

Answer: B
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Re: What is the least integer p for which 27^p > 3^18? &nbs [#permalink] 23 Mar 2018, 12:41
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