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# A tricky question about Exponents

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Manager
Joined: 08 Jun 2011
Posts: 98
Followers: 1

Kudos [?]: 29 [0], given: 65

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26 Jul 2011, 12:31
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Difficulty:

5% (low)

Question Stats:

93% (01:24) correct 7% (00:00) wrong based on 42 sessions

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if m^-1 = -(1/3) then m^-2 equals?

(A) -9
(B) -3
(C) -(1/9)
(D) 1/9
(E) 9

The answer is D and this is how OG12 explained it

m^-2 is (m^-1)^2 = m^-2. Therefore, we square all of -(1/3) which is = 1/9.

This is how I did it.

m^-1 = 1/m^1 Which in turn is 1/m.

This means that m^-2 = 1/m^2 .

So if 1/m = -(1/3) then 1/m^2 should be -(1/3^2) which is -(1/9). Why do I have to square the entire thing up? I am only squaring the bottom hence why would the negative sign go. Perhaps it's for the 1 and not for the 9?
[Reveal] Spoiler: OA
Senior Manager
Joined: 28 Jun 2009
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Location: United States (MA)
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26 Jul 2011, 12:52
1
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Quote:
if 1/m = -(1/3) then 1/m^2 should be -(1/3^2) which is -(1/9).

You're right till
m^-2 = 1/m^2 and m^-1 = 1/m^1 = 1/m

given m^-1 = -(1/3) so, 1/m = -(1/3) solving this, m = -3
Now, m^-2 = 1/m^2 = 1/(-3)^2 = 1/9
Senior Manager
Joined: 03 Mar 2010
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26 Jul 2011, 12:53
See it this way,
$$1/m = -1/3$$
$$m= -3 ( not 3)$$
$$1/m = 1/-3$$
$$1/m^2 = 1/(-3)^2$$
$$1/m^2 = 1/9$$
$$m^-^2 = 1/9.$$
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Manager
Joined: 08 Jun 2011
Posts: 98
Followers: 1

Kudos [?]: 29 [0], given: 65

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26 Jul 2011, 12:59
See it this way,
$$1/m = -1/3$$
$$m= -3 ( not 3)$$
$$1/m = 1/-3$$
$$1/m^2 = 1/(-3)^2$$
$$1/m^2 = 1/9$$
$$m^-^2 = 1/9.$$

So basically you assumed that m=-3 because there was a 1 in the top and 1 cannot possibly = -1 so the m had to have the negative sign, correct?
Senior Manager
Joined: 03 Mar 2010
Posts: 440
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26 Jul 2011, 13:06
Yes. But there is no assumption. It's a fact. It's math.
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