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Re: Manhattan Gmat prep Q [#permalink]
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. _________________

Re: At least one solution [#permalink]
06 Dec 2010, 08:21

2

This post received KUDOS

Expert's post

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. _________________

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

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]

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]

No solution N: Manhattan GMAT test [#permalink]
05 May 2011, 16:52

2

This post was BOOKMARKED

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

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

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.

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