What does the pound sign indicate?

Patrick,

The pound sign is an operator.

When you see an operator, your first thought should be"time to sub-in values."

In this case, we know that "a # b" = -b^4

So, if we have W # Z then we have -Z^4.

And so we know K # L = -L^4.

So what about X # (-Y)? Well that'll be -(-Y^4).

We can simplify.
-(-Y^4) =

-[(-Y) (-Y) (-Y) (-Y)] =

-[(Y) (Y) (Y) (Y)] =
*NOTE*: here we have an even-number exponent (^4), so we know the result MUST be positive. This allows us to wipe out all of the negative signs inside the [brackets]. Another way to look at this step is to pull the four -1's outside the (parentheses) and multiply them together, which would give us a positive 1. Either way you look at it, we get -[(Y) (Y) (Y) (Y)]

So now we have -[(Y) (Y) (Y) (Y)] =

-[(Y)^4] =

-Y^4

Choice (E) wins.

-y^4 different (-y)^4. If a#b is defined by the expresstion a#b=-b^4, so the value of x#(-y) is equal: x# (-1).y <=> x # (-1).(-y^4)<=> x#-y^
I think correct answer is A

Vu Minh Trang, and John. The answer is most definitely not A. Here, "b" is (-y). So, -b^4, which is equal to -1*b^4, expressed in terms of (-y), is -1*(-y)^4. Since a number raises to an even power is the same as its negative raised to the same even power, the expression can be further simplified to -1*y^4, or -y^4.

If the exponent were odd, the answer must be y^4, but we have a positive one so correct answer is E