Each of the following equations has at least one solution EXCEPT

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

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.

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.

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.

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.

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

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

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

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!

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)

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

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

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

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

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.