Multiple Law of exponents

crappiefisher26

New member
Joined
Jul 23, 2006
Messages
49
the original expressions are

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

and
(3x to the 2) to the -3

kinda confused on how to work these type out.
 

Mrspi

Senior Member
Joined
Dec 17, 2005
Messages
2,128
crappiefisher26 said:
the original expressions are

(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.

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

This is a product raised to a power, so we need this rule:

(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 power raised to a power, so we need another rule:

(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.
 

crappiefisher26

New member
Joined
Jul 23, 2006
Messages
49
Im sorry on the first problem the last power should have been 4 not -4
 

jonboy

Full Member
Joined
Jun 8, 2006
Messages
547
Hmm then what do you think the answer would be? Mrspi gave you a magnificent explanation on how to do solve these types of problems (I even learned from that post).
 
Top