1. ## exponents

Can someone please explain how to simplify these, thanks so much.

(2b^4)^-1

(x^2y^-1)^-1

x^2/2x^3

2. Originally Posted by Natalie
Can someone please explain how to simplify these, thanks so much.

(2b^4)^-1

(x^2y^-1)^-1

x^2/2x^3
When you have a term raised to a negative exponent, you take the reciprical of the term.

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

(x^2y^-1)^-1 = 1/(x^2y^-1) = y/x^2 <-- notice that the y had a negative exponent and so could be brought back up the numerator

x^2/(2x^3) <-- in this case, we need to subtract the exponents since the "x"s are the same base. We get
1/2*x^(2 - 3) = 1/2*x^-1 = 1/(2x)

If that's not clear, say so and I'll try to explain it better.

3. Hello, Natalie!

1) .(2b^4)^-1
(2b^4)^-1 .= .(2^-1)(b^4)^-1 .= .(2^-1)·(b^-4)

. . . . . . . . 1
. . . . . . .2b^
4

2) .(x^2·y^-1)^-1
(x^2·y^-1)^-1 .= .(x^2)^-1·(y^-1)^-1 .= .(x^-2)·(y)

. . . . . . . y
. . . . . . .

3) .x²/2x³
. . . . . . . . . . . .1
We have: .---- .= .---
. . . . . . . .2x³ . . .2x