#### crappiefisher26

##### New member

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(2a to the -3) to the -4

and

(3x to the 2) to the -3

kinda confused on how to work these type out.

- Thread starter crappiefisher26
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- Jul 23, 2006

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(2a to the -3) to the -4

and

(3x to the 2) to the -3

kinda confused on how to work these type out.

I'll walk you through the first problem.crappiefisher26 said:

(2a to the -3) to the -4

and

(3x to the 2) to the -3

kinda confused on how to work these type out.

(2 a<sup>-3</sup>)<sup>-4</sup>

This is a

(ab)<sup>n</sup> = a<sup>n</sup> b<sup>n</sup>

When a product is raised to a power, each factor is raised to that power. Apply this rule to your problem:

2<sup>-4</sup> (a<sup>-3</sup>)<sup>-4</sup>

Next, we have a

(a<sup>m</sup>)<sup>n</sup> = a<sup>m*n</sup>

That is, when you raise a power to a power, you multiply the exponents. Apply this rule to your problem:

2<sup>-4</sup> a<sup>(-3)*(-4)</sup>

2<sup>-4</sup> a<sup>12</sup>

Finally, we need to deal with the negative exponent on 2<sup>-4</sup>. Remember that

a<sup>-n</sup> = 1 / a<sup>n</sup>

Apply this rule:

(1 / 2<sup>4</sup>) * a<sup>12</sup>

(1/16) a<sup>12</sup>

or,

a<sup>12</sup> / 16

Now, the second problem is done in much the same way. See what you can do with it. If you have difficulty, please repost showing all of your steps.

- Joined
- Jul 23, 2006

- Messages
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Im sorry on the first problem the last power should have been 4 not -4