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

(A) The left side is always negative, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of this equation are reciprocals when n is odd, and opposite reciprocals when n is even; the absolute values won’t be the same unless n = 0, but the signs won’t be the same unless n is odd. Therefore, the equation has no solution.

(B) The left side is always positive, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of the equation are reciprocals when n is even, and opposite reciprocals when n is odd. The only solution to the equation is n = 0, which produces 1 on both sides.

(C) The left side is always positive, while the right side is positive for even values of n and negative for odd values of n. Therefore, the two sides of the equation are reciprocals when n is even, and opposite reciprocals when n is odd. The only solution to the equation is n = 0, which produces 1 on both sides.

(D) The left side is positive for even values of n and negative for odd values of n, while the right side is always negative; the absolute values of the two sides are always the same (= 2n). Therefore, any odd value of n will solve this equation.

(E) The left side is positive for even values of n and negative for odd values of n, while the right side is always negative; the absolute values of the two sides are always the same (= 2-n). Therefore, any odd value of n will solve this equation.

The correct answer is A.

Why isn't n=0 a viable solution for answer choice (A)? If you plug in n=0, don't you get 1=1?

kung if we pay attention to option A then equation is -
-2^n = (-2)^-n

Here for left hand side the minus sign is not associated with the power of n . hence LHS is always negative.
while the RHS - minus sign is inside bracket to the power of n . RHS is positive for even n , and negative for odd n . hence with n = 0 , the equation becomes
-1 = 1
which is not a possible solution. hence A has no solution at all

ah. parentheses vs. no parentheses, pemdas, etc. thanks.

Remember that when we take powers of negative numbers, parentheses matter.

(-2)^4 = (-2)(-2)(-2)(-2) = 16
- 2^4 = - (2^4) = - (16) = -16

Hope that helps.

_________________
Ben Ku
Instructor
ManhattanGMAT

Hi,

I still don't understand how to tackle this qn nor do I understand the difference between the the different answer options. I was trying to solve this qn by substitution, but I guess that is not the right way. Please let me know the fastest way to tackle such qns.

Regards
Kiran

Hi,

I still don't understand how to tackle this qn nor do I understand the difference between the the different answer options. I was trying to solve this qn by substitution, but I guess that is not the right way. Please let me know the fastest way to tackle such qns.

Regards
Kiran

Hi,

I still don't understand how to tackle this qn nor do I understand the difference between the the different answer options. I was trying to solve this qn by substitution, but I guess that is not the right way. Please let me know the fastest way to tackle such qns.

Regards
Kiran

Kiran,
This is a tough question, which is why it is a 700-800 level question. You have to focus on a few important aspects:
1) pay attention to parentheses, as Ben demonstrated above.
2) make sure you know how to deal with a negative power--for example, know that 5^-2 = 1/25. If you have a negative power you have to take the original number's reciprocal then apply the exponent to the entire new number.

For me the easiest way to do this problem is to think "what are some numbers I can substitute in for n that are easy to work with and may allow me to cross off some answers?" For me what comes to mind is n=0 or n=1. Anything to the 0 power will be 1, whereas anything to the 1 power stays the same. By plugging in 0 for n I can eliminate B and C; by plugging in 1 I can eliminate D and E. Trying plugging those in and writing the equations out. You can probably see why these solutions do work and why those answer choices must be eliminated.

A is the one that does not work for any value of n. On the left side of the equation, we are taking 2^n and then applying a negative sign. That result will always be negative. On the right, I can only get a negative result if I use an odd integer for n. However, my result will be a reciprocal of what is on the left.

For example, if n = 1, I get -2 on the left and -1/2 on the right.
If n=3, I get -8 on the left and -1/8 on the right.
And so on.

Thus, an even n doesn't work because I'll get a negative on the left and a positive on the right. An odd n doesn't work because I get the reciprocal situation described in the paragraph above.

A is the only one without a solution for n.

_________________
Jamie Nelson
ManhattanGMAT Instructor

Why can't n be negative ?

You can certainly test out a negative n if you like, but what's important to focus on is numbers that can help you find answers that work and then make eliminations. 0 and 1 are really good numbers to initially try for n as you can make several eliminations.

_________________
Jamie Nelson
ManhattanGMAT Instructor

is there an algebraic way to do this question?

I don't think so. You don't have much to manipulate, and in very few of these answer choices can you actually manipulate to any extent. Testing obvious values is much quicker and easier.

_________________
Jamie Nelson
ManhattanGMAT Instructor

jnelson your explanation works well. i was stumped when i first encountered this problem on the CAT. but plugging in 0 or 1 when we have variables as exponentials certainly works.

thanks

Thank you! I am very glad to hear that.

_________________
Jamie Nelson
ManhattanGMAT Instructor

I think there's a phrasing problem in this question: it does not indicate that n is an integer. This creates some confusion when I see the explanation considering odd & even cases for n.