Manhattan

[das.ashmita]

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)

OA A

[Whitney Garner]

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)

OA A

Hi das.ashmita!

The best place to start is by re-writing the choices on your paper so we don't confuse the parenthesis:



Okay, so we need to find which of these does NOT have a solution - that means that there is NO value of n that would make this true. Hmmm, that also means that 4 of these WILL be true for at least some value of n. Which would be easier to find?

Well, we could try to simplify/solve these equations by grouping term, etc. But this is the GMAT right, so the easiest thing to do is probably to quickly try a couple of LIKELY numbers to try to eliminate choices. For example, what if N=0? A power of 0 makes any base 1 so that should be easy to work through. Just make sure you're careful with those parenthesis - if the negative is not in parenthesis WITH the base, then you take the exponent FIRST and then apply the sign!



Well, we see that choices B and C are equal when n=0, which means when n=0 they are TRUE, so they have solutions!! We can eliminate them!

Now let's see if we can try another easy number, n=1. When n=1, the base stays the same, but when n=-1 we just put the base under 1 in a fraction. We don't have to test B and C but I will just to be complete!



So in this scenario, choices D and E are equal when n=1, which means they have a solution!! We can eliminate them!

So now we can put this all together - we can eliminate B/C because they are true when n=0, and we can eliminate D/E because they are true when n=1. So that only leaves A!

Hope this helps!

Whit

[Anurag@Gurome]

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)

OA A

(A) -2^n is always negative for all values of n, while (-2)^-n is positive for even values and negative for odd integers.
If n = 0, then -2^n = -1 and (-2)^0 = 1, so for n = 0, it does not hold true.
If n = 1, then -2^n = -2 and (-2)^-n = -1/2, so -2^n is not equal to (-2)^-n and if n = 2, then -2^n = -4 and (-2)^-n = 1/4, again-2^n is not equal to (-2)^-n. This equation gives no solution for any value of n.

(B) If n = 0, then 2^-n = 2^0 =1 and (-2)^n = (-2)^0 = 1. So for n = 0, it holds true.

(C) If n = 0, 2^n = 2^0 = 1 and (-2)^-n = (-2)^0 = 1. So for n = 0, it holds true.

(D) If n = 0, (-2)^n = (-2)^0 = 1 and -2^n = -2^0 = -1. But for any odd value, say n = 1, (-2)^1 = -2 and -2^n = -2^1 = -2. So, (-2)^n = -2^n holds true for odd values.

(E) If n = 0, (-2)^-n = (-2)^0 = 1 and -2^-n = -2^0 = -1. If n = 1, (-2)^-1 = -1/2 and -2^-n = -2^-1 = -1/2. So, (-2)^n = -2^n holds true for odd values.

The correct answer is A.