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# Each of the following equations has at least one solution

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Each of the following equations has at least one solution [#permalink]

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12 May 2010, 09:12
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Each of the following equations has at least one solution EXCEPT

A. –2^n = (–2)^-n
B. 2^-n = (–2)^n
C. 2^n = (–2)^-n
D. (–2)^n = –2^n
E. (–2)^-n = –2^-n
[Reveal] Spoiler: OA
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12 May 2010, 10:53
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marcusaurelius wrote:
Each of the following equations has at least one solution EXCEPT

–2^n = (–2)^-n
2^-n = (–2)^n
2^n = (–2)^-n
(–2)^n = –2^n
(–2)^-n = –2^-n

IMHO A

a) –2^n = (–2)^-n
–2^n = 1/(–2)^n
–2^n * (–2)^n = 1, Keep it. Let's solve the other options..!!

b) 2^-n = (–2)^n
1/2^n = (–2)^n
1 = (–2)^n * (2^n)
For n=0, L.H.S = R.H.S

c) 2^n = (–2)^-n
2^n = 1/ (–2)^n
(2^n) * (–2)^n = 1
For n=0, L.H.S = R.H.S

d) (–2)^n = –2^n
(–2)^n / –2^n = 1
For n=1, L.H.S = R.H.S

e) (–2)^-n = –2^-n
1/ (–2)^n = 1/–2^n
For n=1, L.H.S = R.H.S
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12 May 2010, 11:09
all seems to have n=0 as solution....? whats the OA
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12 May 2010, 14:01
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gurpreetsingh wrote:
all seems to have n=0 as solution....? whats the OA

$$n=0$$ is not a solution of the equation $$-2^n = (-2)^{-n}$$ (in fact this equation has no solution):

$$-2^n=-(2^n)=-(2^{0})=-1$$ but $$(-2)^{-n}=(-2)^{0}=1$$.
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12 May 2010, 15:13
yes right i didnt read it closely
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Re: Manhattan Gmat prep Q [#permalink]

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14 May 2010, 06:56
Hi,

it's really elegant question

0 is a solution for second and third equations.
1 is a solution for the last two equations.

So answer is A. Really if n is not equal to 0 then absolute value of left part is greater than 1 and right part is less than 1. In case when n is equal to 0 we will get -1=1.
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14 May 2010, 12:49
great explanation Bunuel, thanks
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15 Jun 2010, 13:15
Thanks for merging.

Do you mean a negative number raised to the power of 0 yields -1?? I didn't know that!
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15 Jun 2010, 13:32
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study wrote:
Thanks for merging.

Do you mean a negative number raised to the power of 0 yields -1?? I didn't know that!

No.

Any number to the power of zero equals to 1 (except 0^0: 0^0 is undefined for GMAT and not tested).

The point here is that $$-2^n$$ means $$-(2^n)$$ and not $$(-2)^n$$. So for $$n=0$$ --> $$-2^n=-(2^n)=-(2^0)=-(1)$$. But if it were $$(-2)^n$$, then for $$n=0$$ --> $$(-2)^0=1$$.

Hope it's clear.
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15 Jun 2010, 22:56
So how do you know that the point here is that -2^n means -(2^n) and not (-2)^n

The actual question has no parenthesis. This is tricky!
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16 Jun 2010, 05:36
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study wrote:
So how do you know that the point here is that -2^n means -(2^n) and not (-2)^n

The actual question has no parenthesis. This is tricky!

I mean that $$-x^y$$ always means $$-(x^y)$$. If it's supposed to mean $$(-x)^y$$, then it would be represented this way.
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Re: At least one solution [#permalink]

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06 Dec 2010, 08:21
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SubratGmat2011 wrote:
Each of the following equations has at least one solution EXCEPT
-2^n = (-2)^-n
2^-n = (-2)^n
2^n = (-2)^-n
(-2)^n = -2^n
(-2)^-n = -2^-n

Can somebody plz help me out what is the approch for this type of problems?

The first and most straight forward approach that comes to mind is that I can see most of these equations will have n = 0 or n = 1 as a solution.
Except for the very first one:
n = 0: -2^0 = -1 while (-2)^(-0) = 1
n = 1: -2^1 = -2 while (-2)^-n = -1/2

For all other options, n = 0 or 1 satisfies the equation.
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Re: No Solution N [#permalink]

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19 Jan 2011, 22:03
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Lets look at each choice -

A –2^n = (–2)^-n
=>(-1).(2)^n = 1/(-2)^n
=>(-1).(2)^n.(-2)^n = 1
=>(-1).(2)^n.(-1)^n.(2)^n = 1
=>(-1).(-1)^n.(2)^2n = 1
Above cannot be true for any value of n (No solution - answer)

B 2^-n = (–2)^n
=>1/(2)^n = (-2)^n
=>1=(-1)^n.(2)^n.(2)^n
=>1=(-1)^n.(2)^2n
Above is true for n=0, so it has atleast one solution

C 2^n = (–2)^-n
=>(2)^n = 1/(-2)^n
Rest of the steps Similar to option B

D (–2)^n = –2^n
=>(-1)^n. (2)^n = (-1).(2)^n
=>(-1)^n = (-1)
Above is true for all odd values of n

E (–2)^-n = –2^-n
=>1/[(-1)^n. (2)^n] = (-1)/(2)^n
=>1/[(-1)^n] = (-1)
=>1/(-1)^n = -1
Above is true for all odd values of n

I hope this helps.
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20 Jan 2011, 15:19
Bunel, regarding your explanations (2 things you mentioned in your post)
Please correct me if I am miss reading

1. "In fact Option A doesn't have any solutions" - I disagree
LHS: (-2)^n [consider n = 0] then value will be (-2)^0 = 1 [this evaluation related to/depends on 2nd point in my post]
and
RHS: (-2)^-n [consider n = 0]then value will be (-2)^-0=(-2)^0= 1

2. "LSH actually should be read as -1 * (2)^n"
Does the rule says that if parenthesis are missing then always start with "power" first and then assign the -ve or +ve signs to the calculated number ? in that case I agree i.e it should be read as -2^n is to be read as -1[assign this last]* 2^n[solve this 1st]

Please confirm, as always thanks for help
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20 Jan 2011, 15:34
pushkarajg wrote:
Bunel, regarding your explanations (2 things you mentioned in your post)
Please correct me if I am miss reading

1. "In fact Option A doesn't have any solutions" - I disagree
LHS: (-2)^n [consider n = 0] then value will be (-2)^0 = 1 [this evaluation related to/depends on 2nd point in my post]
and
RHS: (-2)^-n [consider n = 0]then value will be (-2)^-0=(-2)^0= 1

2. "LSH actually should be read as -1 * (2)^n"
Does the rule says that if parenthesis are missing then always start with "power" first and then assign the -ve or +ve signs to the calculated number ? in that case I agree i.e it should be read as -2^n is to be read as -1[assign this last]* 2^n[solve this 1st]

Please confirm, as always thanks for help

Your doubts are answered on previous page:
exponents-94119.html#p724081
exponents-94119.html#p738365
exponents-94119.html#p738571
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18 Feb 2011, 10:22
i guess the only real way to solve it under 2 min is to plug in 0/1...

if u start with choosing 1 here as first step is not good... choosing 0 is canceling 3 choices quickly...
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No solution N: Manhattan GMAT test [#permalink]

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05 May 2011, 16:52
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I am totally lost on this one ..... can anyone help please ?

Each of the following equations has at least one solution EXCEPT
A.) –2^n = (–2)^-n
B.) 2^-n = (–2)^n
C.) 2^n = (–2)^-n
D.) (–2)^n = –2^n
E.) (–2)^-n = -2^-n
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Re: No solution N: Manhattan GMAT test [#permalink]

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05 May 2011, 17:26
lets pick a value for n.

n = 0

A. cannot be true as we get -1 on LHS and 1 on RHS ( as anything to the power of 0 is 1)
B. true (LHS = RHS = 1)
C. true (LHS = RHS = 1)
D. true (LHS = RHS = 1)
E. true (LHS = RHS = 1)

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Re: No solution N: Manhattan GMAT test [#permalink]

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05 May 2011, 22:42
for B and C n = 0
for D and E n = 1.

A prevails.
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Re: Each of the following has at least one solution EXCEPT [#permalink]

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18 Aug 2012, 15:14
arichardson26 wrote:
Each of the following has at least one solution EXCEPT

A. -2^n = (-2)^-n

B. 2^-n = (-2)^n

C. 2^n = (-2)^-n

D. (-2)^n = -2^n

E. (-2)^-n = -2^-n

[Reveal] Spoiler:
A

B, C have can be equated by using n=0
D and E have external/independent -ve signs, so 0 wont help, but using n= +1 for D and -1 for E will equate the sides.

Took more than 2 mins
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Re: Each of the following has at least one solution EXCEPT   [#permalink] 18 Aug 2012, 15:14

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