If a and b are nonzero integers, which of the following...

[cschmidlapp]

If a and b are nonzero integers, which of the following must be negative?

a) (-a)^(-2b)
b) (-a)^(-3b)
c) -(a^(-2b))
d) -(a^(-3b))
e) None of these

I understand the logic and the explanation around a being negative or positive, but less so with b. Does b's sign even matter because a taken to a negative power translate to (1/a(^exponent))?

Can someone please explain mechanically why each of the answers is wrong and why C is right? Thank you.

[jnelson0612]

If a and b are nonzero integers, which of the following must be negative?

a) (-a)^(-2b)
b) (-a)^(-3b)
c) -(a^(-2b))
d) -(a^(-3b))
e) None of these

I understand the logic and the explanation around a being negative or positive, but less so with b. Does b's sign even matter because a taken to a negative power translate to (1/a(^exponent))?

Can someone please explain mechanically why each of the answers is wrong and why C is right? Thank you.

Sure, let me assign values and show you why A, B, and D could be positive. Then we'll discuss why C must be negative.

a) (-a)^(-2b)
a=-2
b=-2
That gives me (2)^(4). Positive.

b) (-a)^(-3b)
a=-2
b=-2
That gives me (2)^(6). Positive.

d) -(a^(-3b))
This is tougher. I have a negative sign on the outside of all of the parentheses, so I want to make the inside negative. Two negatives create a positive. Let me use:
a=-1
b=-1
-(-1^(3)). That is -(-1) or 1. Positive.

Now, let's look at why C *must* be negative:
c) -(a^(-2b))

Right away, I notice that the exponent is -2b. Anytime I multiply an integer (b) by an even number (-2) I get an even product. So -2b will be even.

So on the inside I have (a^even integer). An even integer always creates a positive result, whether the base is positive or negative (assuming that the base is not zero, and in this case it cannot be). Thus (a^even integer) is a positive number.

Finally, the full expression is -(positive number). This must be a negative result and this is why C MUST be negative.

[cschmidlapp]

Ok, so a^(even integer) will always be positive, even if the even integer exponent is negative? Or do you have to make it positive by multiplying by -b?

Thanks.

[jnelson0612]

Ok, so a^(even integer) will always be positive, even if the even integer exponent is negative? Or do you have to make it positive by multiplying by -b?

Thanks.

Yes, and let's discuss why. A negative exponent means that you need to do two things:
1) take the reciprocal of the base. That is all we need to do to deal with the negative component of the exponent.
THEN
2) apply the positive exponent.

Thus, if I have 3^-2, that turns into (1/3)^2.

Even if I have a negative base, that negative even exponent will simply flip the base over then make the base positive through the application of the even exponent. For instance:
-4^(-2) = (1/4)^2 = 1/16.

I hope that this makes more sense, but please let us know if you need further help. :-)