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

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Math Expert
Joined: 02 Sep 2009
Posts: 52164

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02 Mar 2017, 07:05
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Difficulty:

15% (low)

Question Stats:

82% (00:37) correct 18% (00:59) wrong based on 71 sessions

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

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

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02 Mar 2017, 07:41
1
Bunuel wrote:
$$\frac{9^{(-4)}}{27^{(-3)}} =$$

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

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

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02 Mar 2017, 07:49
Inverting the signs.

Fraction becomes $$\frac{27^3}{9^4}$$ = $$\frac{3^9}{3^8}$$ = 3
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02 Mar 2017, 08:13
$$\frac{3^{-8}}{3^{-9}} = 3^{-8-(-9)} = 3^{1} = 3$$
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02 Mar 2017, 11:40
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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02 Mar 2017, 13:34
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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02 Mar 2017, 13:38
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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03 Mar 2017, 00:50
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
Re: 9^(-4)/27^(-3) = &nbs [#permalink] 03 Mar 2017, 00:50
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