HELP ASAP! Need help with multiplying variable exponants...?

Question

I already have the answer. I just need to see it worked out so I can figure it out. Thanks!

(3^-2)^a / (3^-2)^b * (3^b)^-2a / (9^-b)^-3a

The answer is : 3^-2a+2b-8ab

I have no idea how the answer came to fruitition in the answer book... I need to see it worked out, please!

Answer 1

Providing extra brackets, I assume the problem is properly stated as:

[(3^-2)^a / (3^-2)^b] * [(3^b)^-2a / (9^-b)^-3a]

Two fractions multiplied together. You need to remember some rules for exponents:

1) (x^a)^b = x^ab
2) x^1/x^b = x^(a - b)
3) x^a(x^b) = x^(a + b)

Applying rule 1

[3^-2a/3^-2b] * [3^-2ab/9^3ab]

Since 9 = 3^2, the denominator of the right fraction can be written

(3^2)^3ab or 3^6ab

Now apply rule 2

3^(-2a + 2b) * 3^(-2ab - 6ab)

And rule 3

3^(-2a +2b - 2ab - 6ab)

or

3^(-2a + 2b - 8ab)

Again, it is important to use brackets to make your work absolutely clear. In other words, the answer you gave could be interpreted as

(3^-2a) + 2b - 6ab

Answer 2

Ok, lets start by breaking this into pieces and handling the bracketed pieces first.

(3^-2)^a - When you have something that is raised to a power, raised to another power, you can multiply the 2 powers together. So this piece becomes: 3^-2a
The other brackets then become
(3^-2)^b = 3^-2b
(3^b)^-2a = 3^-2ab
(9^-b)^-3a = 9^3ab (careful, the negatives cancel themselves out here!

Ok, now lets rewrite it without the brackets
3^-2a/3^-2b*3^-2ab/9^3ab

Then next thing to do is to rewrite this as a multiplication problem, with the division parts as being under 1. Like so:

(3^-2a) * (1/(3^-2b)) * (3^-2ab) * (1/(9^3ab)

To simplify further, lets get rid of the division portions. Whenever you would like to take a denominator which has a power in it, and make it into a numerator, all you have to do is change the sign of the power. So we'll take the 2 denominators and make them numerators, like so:

1/(3^-2b) = 3^2b
1/(9^3ab) = 9^-3ab

And we rewrite our problem:

3^-2a * 3^2b * 3^-2ab * 9^-3ab

Now we're close, but there is that one part of th problem that has a 9 in it. That is no good. So in order to make that look like the others, you just need to think of 9 in terms of 3s. 9 is 3 raise to the power of 2, so that portion of the equation can be rewritten as: (3^2)^-3ab
Just like before, we can multiply powers and so we get: 3^-6ab

And we rewrite our formula:

3^-2a * 3^2b * 3^-2ab * 3^-6ab

And finally, whenever we have a multiplication problem, in which we are multiplying powers that have the same base, we can just add the powers together attached to the single base. So we add the powers to get:

3^(-2a + 2b + -2ab + -6ab)

We combine our ab terms and get your final answer:

3^(-2a + 2b - 8ab)

Answer 3

First simplify the exponants and you´ll get

3^-2a/3^-2b * 3^-2ab/ 9^3ab

The first part you can resolve because the base is the same. So, since it´s a division you keap the base and subtract the exponants and get

3^(-2a+2b) * 3^-2ab/9^3ab

In the second part you don`t have neither the same base nor the same exponant. But 9= 3^2. So, you can substitute and get

3^(-2a+2b) * 3^-2ab/ 3^(2*3ab)

Now, you have the same base. So you can subtract the exponants

3^(-2a+2b) * 3^(-2ab-6ab)

So, again, same base but, now, since you have a multiplication, you have to add the exponants

3^(-2a+2b-8ab)

Problem solved

Answer 4

((3^-2)^a)/((3^-2)^b) * ((3^b)^(-2a))/((9^-b)^(-3a))
((3^(-2a))/(3^(-2b)) * ((3^(-2ab))/(9^(3ab))
(3^(-2a + -2ab))/(3^(-2b) * 9^(3ab))
(3^(-2a(1 + b)))/(3^(-2b) + (3^2)^(3ab)))
(3^(-2a(1 + b)))/(3^(-2b) * 3^(6ab))
(3^(-2a(1 + b)))/(3^(-2b + 6ab))
(3^(-2a(1 + b)))/(3^(-2b(1 - 3a)))
((3^-2)^(a(1 + b)))/((3^-2)^(b(1 - 3a)))
((3^2)^(b(1 - 3a)))/((3^2)^(a(1 + b)))
(9^(b(1 - 3a)))/(9^(a(1 + b)))

9^(b(1 - 3a) - a(1 + b))
9^(b - 3ab - a - ab)
9^(b - a - 4ab)

9^(b - a - 4ab) =
(3^2)^(b - a - 4ab) =
3^(2(b - a - 4ab)) =
3^(2b - 2a - 8ab)

this is the same as saying 3^(-2a + 2b - 8ab)

I hope you understood what i did.

If you are wondering why

((3^-2)^(a(1 + b)))/((3^-2)^(b(1 - 3a)))
became
((3^2)^(b(1 - 3a)))/((3^2)^(a(1 + b)))

is because of the negative exponent, since both the denominator and numerator have negative exponents, the fraction gets flipped.

Answer 5

ewww math

Answer 6

umm.yes.....hm....ahh haaa!!!!! Wait..no still nothing. eww math.