\(-2^{-2}\)

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

A should be [-2]^-2 = 2^-2 = 1/4. ANS (C), right? Because [-2] = 2, isn't it?

Ok let me clear the air, if possible, once for all

QA

\([-2]^{ 10y-4x }\)

QB

\(2^{2(5y-2x)}\)

so

\(2^{10y-4x}\)


Now, the main equation multiplied by -2 will become \(-4x+10y=-2\)

Substitute the -2 in QA and QB and we have

QA \(-2^{-2}=-\frac{1}{2^2}=-\frac{1}{4}\)

QB \(2^{-2}=\frac{1}{2^2}=\frac{1}{4}\)

B is the answer