Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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. _________________

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 \(n=0\) --> \((-2)^0=1\).

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.

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...