Help with simplifying expressions

Can someone please show me how to solve these two problems? Thanks!

1) (4)^-5/2

2) (2^-1 / 2^-2)^2

Negative exponents just mean that you put the term, with positive exponent, under 1: it's like a reciprocal. For example:

x^-1 would be 1/x

With fraction exponents, the numerator is the exponent and the denominator is the degree of the root. So x^(1/2) would be sqroot(x).

So, for (1):
I assume you mean -5/2 is your exponent.
1. Get rid of the negative exponent by taking the reciprocal:
1/[4^(5/2)]
2. Take the square root of 4^5, which would be (4^2)*sqrt(4) or 16*sqrt(4), then simplified to 16*2 = 32
3. Remember that it's all under 1, so the answer would be 1/32

(2)
[(2^-1) / (2^-2)]^2
1. You can cancel exponents on top and bottom, so it would become:
[1/(2^-1)]^2
2. Then get rid of the negative exponent:
[1/(1/2)]^2
3. Then combine into one fraction by multiplying reciprocals:
1 * 2/1 = 2
4. Then square it: 2^2 = 4

1. Alternately you can get rid of the negative exponents first by taking reciprocals then recombine into a single fraction:
[(1/2)/(1/4)]^2 = (1/2 * 4/1) = 4/2 = 2
2. Then square it: 2^2 = 4.

Hope this helps :-) This kind of problem is easy once you learn the laws of exponents.

Edit: One good resource is wolframalpha.com, which lets you enter equations and gives you the answers. Don't use it just to get answers, but for verification after doing the steps yourself, and especially when asking how to solve things online (just to make sure people are correct :-).

For the first one using indices=1/(squart of 4)^5 and the second quest =1/2 divided by (1/4)^2=8

1) 4^-5/2 = 1/(4^5/2) = 1/(sqrt(4^5)) = 1/sqrt(1024) = 1/32

2) (2^-1 / 2^-2)^2 = (2^2/2^1)^2 = (4/2)^2 = 2^2 = 4

1) (4)^-5/2= (2^2)^-5/2=(2)^-5=1/32

2) (2^-1/2^-2)^2=(2^-4)^2=2^-8=1/256