\(-2^{-2}\)

will be

\(-\frac{1}{4}\)

see more on exponents https://gre.myprepclub.com/forum/gre-ma ... 24948.html

Carcass

if we are -2^2 does that means the negative will remain there even if we have even power ??

in the question, the negative is inside the brackets.

My solution is as follows:

I first simplified
quantity A :
\(\frac{((-2)^10y)}{(-2^4x)}\)

Quantity B :
I simplified 4
\frac{[fraction]2^(10y)}{2^(4x)}[/fraction]

I think the answer is C

rule of exponent


\(x^{-n}=\frac{1}{x^n}\)

\(7^{-2}=\frac{1}{7^2}\)

so having \(-7^{-2}= - \frac{1}{7^2}=-\frac{1}{49}\)

I am interpreting the question this way \((-)(2)\)

Otherwise the answer would be C.

The OA reports B so should be as above

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.