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# (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)

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Manager
Joined: 02 Dec 2012
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29 Dec 2012, 05:58
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5% (low)

Question Stats:

83% (00:59) correct 17% (01:25) wrong based on 1865 sessions

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$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) $$(\frac{1}{2})^{(-48)}$$

(B) $$(\frac{1}{2})^{(-11)}$$

(C) $$(\frac{1}{2})^{(-6)}$$

(D) $$(\frac{1}{8})^{(-11)}$$

(E) $$(\frac{1}{8})^{(-6)}$$
[Reveal] Spoiler: OA
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29 Dec 2012, 06:01
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$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=2^3*4^2*16=2^3*2^4*2^4=2^{11}=(\frac{1}{2})^{-11}$$.

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29 Dec 2012, 06:14
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$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

whenevr you see a fraction raised to a negative power think reciprocal

so we have : 2^3 * 4^2 * 16^1

Consolidate: 2^3 * 2^4 * 2^4

Clearly as you can see we have 2^11 only B fits

otherwise take the reciprocal of 2^11 ----> 1/2^-11

B is the best
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21 Mar 2014, 00:21
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Keep the base as 2 & modify the powers accordingly

$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}$$

= $$(\frac{1}{2})^{-3}(\frac{1}{2})^{-4}(\frac{1}{2})^{-4}$$

With the base the same, add the powers

-3-4-4 = -11

$$(\frac{1}{2})^{-11}$$
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03 Dec 2014, 07:20
Resolved it as 16 x 2^3 x 4^2
After 2:20sec
Simplified it as 1/2
B
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15 Jul 2016, 04:41
$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

We start by using the negative exponent rule. When a fractional base is raised to a negative exponent, we can rewrite the expression (without the negative exponent) by flipping the fraction and making the exponent positive. For example, (1/2)^-3 = 2^3

We are using the negative exponent rule because it’s not only easier to deal with positive exponents, but also when we flip the fractional base, the fraction becomes an integer.

(1/2)^-3 = 2^3

(1/4)^-2 = 4^2 = (2 x 2)^2 = (2^2)^2 = 2^4

(1/16)^-1 = 16^1 = (2 x 2 x 2 x 2)^1 = (2^4)^1 = 2^4

We multiply each term in the expression, obtaining:

2^3 x 2^4 x 2^4

Remember, since the bases are the same, we keep the base and add the exponents. We are left with:

2^(3+4+4) = 2^11

Finally, since our answer choices are expressed in fractional form, we once again have to use the negative exponent rule. To convert a base with a positive exponent, take the reciprocal of the base and change the positive exponent to its negative counterpart. Using the rule we get:

2^11 = (1/2)^-11

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30 Nov 2016, 17:48
Expert's post
Top Contributor
$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

Nice rule: $$(\frac{a}{b})^{-n} = (\frac{b}{a})^{n}$$

$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1} = (\frac{2}{1})^{3}(\frac{4}{1})^{2}(\frac{16}{1})^{1}$$

$$= (2^{3})(4^{2})(16^{1})$$

$$= (2^{3})(2^{4})(2^{4})$$

$$= 2^{11}$$

$$= (2/1)^{11}$$

$$= (1/2)^{-11}$$

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07 Dec 2016, 17:28
(2)^3 (4)^2 (16)^1
Simplify to base 2
(2)^3 (2)^4 (2)^4
2^11
(1/2)^-11
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Manager
Joined: 30 Apr 2013
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22 Aug 2017, 04:16
I understood the method but I am just wondering if we could do this way - i tried and got the wrong answer

(1/2)^(-3)* (1/4)^(-2)*(1/16)^(-1)

= 1/(1/2)^3 * 1/(1/4)^2* 1/(1/16)^1

=1/(1/8)*1/(1/16)*1/(1/16)
=8/1*16/1*16/1

Please let me know why cant we do this way?
Math Expert
Joined: 02 Sep 2009
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22 Aug 2017, 04:21
santro789 wrote:
I understood the method but I am just wondering if we could do this way - i tried and got the wrong answer

(1/2)^(-3)* (1/4)^(-2)*(1/16)^(-1)

= 1/(1/2)^3 * 1/(1/4)^2* 1/(1/16)^1

=1/(1/8)*1/(1/16)*1/(1/16)
=8/1*16/1*16/1

Please let me know why cant we do this way?

You can and you'll get the same answer: 8/1*16/1*16/1 = 2^11 = (1/2)^(-11)
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24 Aug 2017, 21:39
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Hello Moderators,

Do you think it would be good to make the answer choices math friendly - just like the main stem? This is OG question and it would be good to see as it would appear in the real exam.

$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

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Joined: 02 Sep 2009
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24 Aug 2017, 21:44
susheelh wrote:
Hello Moderators,

Do you think it would be good to make the answer choices math friendly - just like the main stem? This is OG question and it would be good to see as it would appear in the real exam.

$$(\frac{1}{2})^{-3}(\frac{1}{4})^{-2}(\frac{1}{16})^{-1}=$$

(A) (1/2)^(-48)
(B) (1/2)^(-11)
(C) (1/2)^(-6)
(D) (1/8)^(-11)
(E) (1/8)^(-6)

_______________
Edited. Thank you.
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Re: (1/2)^(-3)*(1/4)^(-2)*(1/16)^(-1)   [#permalink] 24 Aug 2017, 21:44
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