Each of the following equations has at least one solution EXCEPT

Question

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}

Bunuel · Accepted Answer

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 .

nverma · Answer

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

Bunuel · Answer

Each of the following equations has at least one solution EXCEPT

A.  –2^n = (–2)^{-n}

If  n = 0 , then  –2^n = -2^0=-1  and  (–2)^{-n}=(–2)^{-0}=1 .

B.  2^{-n} = (–2)^n

If  n = 0 , then  2^{-n} = 2^{-0}=1  and  (–2)^n=(–2)^0=1

C.  2^n = (–2)^{-n}

If  n = 0 , then  2^n = 2^0=1  and  (–2)^{-n}=(–2)^{-0}=1

D.  (–2)^n = –2^n

If  n = 0 , then  (–2)^n = (–2)^0 = 1  and  –2^n=–2^0=-1

E.  (–2)^{-n} = –2^{-n}

If  n = 0 , then  (–2)^{-n} =(–2)^{-0} =1  and  –2^{-n}=–2^{-0}=-1

Hope it helps.

gurpreetsingh · Answer

all seems to have n=0 as solution....? whats the OA

study · Answer

Thanks for merging.

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

Bunuel · Answer

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.

study · Answer

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!

Bunuel · Answer

I mean that  -x^y   always  means  -(x^y) . If it's supposed to mean  (-x)^y , then it would be represented this way.

KarishmaB · Answer

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.

ravsg · Answer

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

I hope this helps.

Legendaddy · Answer

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

NikRu · Answer

Thank you for your response.

I would like to double check why we say that n=0 could be a solution in case of  (-2)^{-n} 
as  (-2)^{-n} = (-2)^{1/n}  and then we can not divide by zero?

Nik

KarishmaB · Answer

(-2)^{-n} = 1/(-2)^n  not  (-2)^{1/n}

jlgdr · Answer

Why did you plug n=1 for the last two, wouldn't it be easier just to plug n=0 for all and see that A has no solution? Just want to know if there was any specific reason why you did so Thank you Cheers J

PS. Would be nice if we could get this question in code format!

KarishmaB · Answer

We need to find the equation that has no solution. What we are trying to do is find at least one solution for 4 equations. The fifth one will obviously not have any solution and will be our answer.
Options (D) and (E) do not have 0 as a solution.
So you try n = 1 on (A), (D) and (E). 
n = 1 is still not a solution for (A) but it is for (D) and (E).

(D) (–2)^n = –2^n
When you put n = 0, you get
(-2)^0 = -2^0
1 = -1 which doesn't hold. 
So you try n = 1
 (–2)^1 = -2^1
-2 = -2
n = 1 is a solution.

Same logic for (E)

hudhudaa · Answer

How do you determine that we should use 0 or 1? I used 2 and 3 and found all of them to not equate.

KarishmaB · Answer

0, 1, and -1 are special cases. Usually we start with them in case we are looking at plugging numbers.

BrentGMATPrepNow · Answer

Rather than trying to solve each individual equation, it will be much faster to recognize that some POSSIBLE solutions include n = 0, n = 1 and n = -1
We should be able to quickly eliminate answer choices by testing possible solutions.

Let's start with n = 0
We get: 
A.  –2^0 = (–2)^{-0}  becomes  -1 = 1 . Doesn't work. Keep answer choice A for now.

B.  2^{-0} = (–2)^0  becomes  1 = 1 . Works!. Eliminate B.

C.  2^0 = (–2)^{-0}  becomes  1 = 1 . Works!. Eliminate C.

D.  (–2)^0 = –2^0  becomes  1 = -1 . Doesn't work. Keep answer choice D for now

E.  (–2)^{-0} = –2^{-0}  becomes  1 = -1 . Doesn't work. Keep answer choice E for now

Now let's try n = 1 with the remaining three answer choices
We get: 
A.  –2^1 = (–2)^{-1}  becomes  -2 = -\frac{1}{2} . Doesn't work. Keep answer choice A for now.

D.  (–2)^1 = –2^1  becomes  -2 = -2 . Works!. Eliminate D.

E.  (–2)^{-1} = –2^{-1}  becomes  -\frac{1}{2} = -\frac{1}{2} . Works!. Eliminate E

By the process of elimination, the correct answer is A

apurv09 · Answer

can someone explain when taking n=0, what happens in option D and E

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