Re

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- May 6th 2007, 08:50 AM #1

- Joined
- Apr 2007
- Posts
- 2

## Some help please

this is just practice problems that I am trying to get a hang of:

a ^-4 b ^3 (a ^7 b ^5)

____________________________

x ^-9 y (x ^9 y ^3)

now the choices I have to choose from are:

a ^3 b ^8

_________

y ^3

a ^3 b ^8

_________

xy ^4

a ^3 b ^8

_________

y ^4

a ^0 b ^10

__________

y ^4

none of these

I need help in learning how to do this type of problem.. Thanks

- May 6th 2007, 08:55 AM #2

- May 6th 2007, 11:01 AM #3
everything is being multiplied in the top and bottom right? ok, here's how we deal with this.

the following are laws of exponents that you should know (you sjould know more than these, but these are the ones you need for this problem).

1) when we multiply two numbers of the same base, we add the powers. that is,

(x^n)*(x^m) = x^(n+m)

example: (x^2)*(x^3) = x^(2+3) = x^5

2) when we divide two numbers of the same base, we subtract the power of the lower one from the power of the top one, that is:

(x^n)/(x^m) = x^(n-m)

example: (x^5)/(x^2) = x^(5-2) = x^3

don't worry if the top power is smaller than the bottom one, still subtract

example 2: (x^2)/(x^7) = x^(2-7) = x^-5

3) the last example should put a question into your head...well, not really, but yeah. How do we deal with negative powers? what does it mean to have a negative power?**negative powers do not change the sign of a number,**when we have a negative power, it means we take the inverse on the number, it means put 1 over the number with a positive power.

example: x^-n = 1/(x^n)

example 2: x^-5 = 1/(x^5)

example 3: (x/y)^-3 = (y/x)^3 = (y^3)/(x^3)

4) Anything (other than zero) raised to the zero power is 1

example: x^0 = 1,

35^0 = 1,

67394.678924^0 = 1

i think that's all we need to tackle this problem, so now let's see how qbkr21 got to his answer.