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Each of the following equations has at least one solution [#permalink]
 12 May 2010, 10:12
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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
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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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This was difficult for me because I didn't understand the significance of the parentheses. I am still having trouble seeing why D is incorrect.
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Re: Manhattan Gmat prep Q [#permalink]
14 May 2010, 07:56
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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great explanation Bunuel, thanks
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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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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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Re: At least one solution [#permalink]
06 Dec 2010, 09:21
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 as a solution since 2/-2 to the power 0/-0 is 1. Except for the very first one: -2^0 = -1 while (-2)^(-0) = 1 For all other options, n = 0 satisfies the equation.
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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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pushkarajg wrote: Can someone please help me with this question, I am really not convince with OA
Each of the following equations has at least one solution EXCEPT
A –2^n = (–2)^-n , n=0 => -1 != 1; solution
B 2^-n = (–2)^n , n=0 => 1 = 1;
C 2^n = (–2)^-n , n=0 => 1 = 1;
D (–2)^n = –2^n , n=0,1;
E (–2)^-n = –2^-n ,n= 0,1;
I believe this helps.
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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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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#p724081exponents-94119.html#p738365exponents-94119.html#p738571
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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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Re: Each of the following has at least one solution EXCEPT [#permalink]
18 Aug 2012, 16:14
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 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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Each of the following has at least one solution EXCEPT [#permalink]
23 Aug 2012, 04:45
OA has to be A because Equation 1 simplifies to (2)^n (2)^n (-1)^n= -1 has no solution for any value of n Rest of options have at least 1 solution
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Each of the following has at least one solution EXCEPT
[#permalink]
23 Aug 2012, 04:45
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