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9^(-4)/27^(-3) =

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9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 08:05
00:00
A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

75% (00:29) correct 25% (00:44) wrong based on 71 sessions

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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 08:41
1
Bunuel wrote:
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27


\(\frac{9^{-4}}{27^{-3}}=9^{-4}/27^{-3} = \frac{1}{9^4} / \frac{1}{27^3}=\frac{27^3}{9^4}=\frac{3^9}{3^8}=3\)

The answer is C
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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 08:49
Inverting the signs.

Fraction becomes \(\frac{27^3}{9^4}\) = \(\frac{3^9}{3^8}\) = 3
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9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 09:13
\(\frac{3^{-8}}{3^{-9}} = 3^{-8-(-9)} = 3^{1} = 3\)
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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 12:40
Bunuel wrote:
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27


\(\frac{9^{(-4)}}{27^{(-3)}}\)

\(= \frac{3^{(-8)}}{3^{(-9)}}\)

\(= 3^{( 9 - 8)}\)

\(= 3^1\)

Thus, answer must be (C) 3
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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 14:34
At a quick glance of the question you can easily eliminate A,D, & E because they are repeats from the question/too easily derived. That leaves B & C, inverting the fraction leaves exponents 9-8=1/base remains 3. 3^1. Answer=C.
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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 02 Mar 2017, 14:38
Bunuel wrote:
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27



it is obviously 3. the answer is C.

(3^-8)/(3^-9)=3^1=3
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Re: 9^(-4)/27^(-3) =  [#permalink]

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New post 03 Mar 2017, 01:50
Bunuel wrote:
\(\frac{9^{(-4)}}{27^{(-3)}} =\)

A. 1/3
B. 1
C. 3
D. 9
E. 27


IMO C
the problem can be simplified as (27*27*27)/(9*9*9*9)= (3*3*3)/(9) = 3
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Re: 9^(-4)/27^(-3) = &nbs [#permalink] 03 Mar 2017, 01:50
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