The main confusion here is because the brackets are not specified in Quantity B. I fell into this trap too.
Quantity B is supposed to be: -(2^2x) and not (-2)^2x. This makes a huge difference, because irrespective of x, this quantity is going to remain negative.

The main counter argument to those arguing that the absence of brackets introduces a lot of ambiguity is that Quantity A DOES USE BRACKETS. (-3)^x is shown explicitly. Which means if they intended Quantity B to be (-2)^2x, they would have specified the brackets.

I think the answer is B as x is odd integar..and also i couldnot understand the power for -2 is (2X) or only X

x is a positive, odd integer.

Quantity A: (-3)^x
Quantity B: -2^(2x)

Note the difference in the 2 expressions:

Qty A: the power of x is applicable on (-3)
Qty B: the power of 2x is applicable on 2 and then the result is negated

Since x is odd: (-3) is raised to an odd power
=> [(-1)*(3)] is raised to an odd power

Thus, $(-1)^x = -1$ and 3^x remains as it is

Hence: Quantity A: $-3^x$

Also, Quantity B: -2^(2x) = -1 * 2^(2x) = -1 * (2^2)^x = -1 * 4^x = $- 4^x$

Thus, we now need to compare: $- 3^x$ and $- 4^x$


Clearly, $3^x < 4^x$ since x is positive integer

=> $- 3^x > - 4^x$ (inequality reverses on multiplying a negative)


Answer A

ans is a because in qt.b is -1*some value by substituting value of x we will get some value which is multiplied by -1 later on

This is a common mistake which even I was about to make. So nope because -2^(2x) becomes -4^x and since x will always be a positive odd integer option B will always be greater.

I plugged 3, and then 1. Both times A > B, and because the plug-in numbers are narrowed down to positive odd numbers, there is no change in this trend.

(-3)^3 = -27 > -(2^2(3)) = -64

(-3)^1 = -3 > (-2)^2 = -4

In the case of Quantity A, $(-3)$ is raised to the power of $x$, which is positive and odd. Since the odd powers preserve the sign of the base, Quantity A will always be negative

Let us take both conflicting cases

In the case of Quantity B, $2$ is raised to the power $2x$, and then the result is negated by the negative sign outside the bracket $-(2^{2x})$. In this case the smaller of two adjacent numbers when raised to twice the power of the larger one, will be much bigger than the other one. That is $2^{2x}$ will be many times greater than $3^x$. If this isn't obvious, you can test it by letting $x=1,3$ and $5$ and check. Thus, Quantity A is greater since we are computing $-(2^{2x})$ and $(-3)^x$.

OR

In the case of Quantity B, $2$ is raised to the power $2$, and the result $4$ is raised to the power $x$. In this case it is obvious that between $(-3)^x$ and $-(4^x)$, the latter will be smaller than the former, so Quantity A is the greatest.

So whatever is your interpretation of $2^{2x}$, the answer is Quantity A.

No. Its -(2^2x) and not (-2)^2x.

Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.