# Homework Help: Exponent questions

1. Aug 2, 2011

### Nelo

1. The problem statement, all variables and given/known data

http://i51.tinypic.com/2zyfyaa.jpg

Question 3 l )

2. Relevant equations

3. The attempt at a solution

(6a^-2b^-3)^2
(__________) 1/6a^2 * 1/b^3 / 1/2a^2 * 1/b^1
(2a^2b^-1 )

I multiplied the top half together to give me 1/ 36a^4b^2

Multiplied the bottom half to give me 1/4a^4b^1 , Then took the reciprical of the bottom half and move it up and multiplied it with the top half.

Giving me a final of

:: 4a^4b^1
__________
36^a4b^5

However, when i fold this down it becomes 1/9 and the a's cancel and 4b is left over at the bottom.

The answer is a^8b^4 / 9

I know its complicated to read, just write it on paper and ull understand.

What did i do wrong?

2. Aug 2, 2011

3. Aug 2, 2011

### Nelo

Or... f) (-3m^-3n^-1)^-3

Solving this one, I get -27m^9 / n^3

Why in the answer book is it m^9*m^3 /27 ?

Anyone....?

4. Aug 2, 2011

### eumyang

You continue to write out your problems in a way that makes it hard to read. I think others would be less willing to help you because of this. Learn LaTeX, for goodness sake!

In any event, if I am reading your work correctly, this line:
is wrong. It should be 1/ 36a^4b^6.

Anyway, your approach is rather confusing. It looks like you want to start by rewriting the negative exponents to positive ones. I wouldn't do that. Instead, divide the numerator by the denominator, subtracting exponents as you go:
$\left(\frac{6a^{-2}b^{-3}}{2a^2 b^{-1}}\right)^{-2} = \left(3a^{-4}b^{-2} \right)^{-2}$
Then "distribute" the -2 exponent that is outside the parentheses, and THEN rewrite any negative exponents that are remaining to positive ones.

5. Aug 2, 2011

### Nelo

If i do it your way, then I get no fractions in my final answer. i just get 9a^8b^4, Sorry I dont know any other method except this recipricol one and its just not working.

I understand your step, but what step comes after How do i turn that into a fraction again

6. Aug 2, 2011

### eumyang

You got it wrong because it looks like you think that
$(-3)^{-3} = -27$,
and it's not. It should be
$(-3)^{-3} = -\frac{1}{27}$.

Also, the answer book does not say that! Don't you see the typos? It should be
$$-\frac{m^9 n^3}{27}$$

7. Aug 2, 2011

### eumyang

No, you should have gotten
$\frac{1}{9}a^8 b^4 = \frac{a^8 b^4}{9}$.
You're not applying the negative exponent to the coefficient correctly, it seems.

8. Aug 2, 2011

### Nelo

Cept i do realise that it is 1/-27, and yes it is a typo, but my question remains the same. Why is that mn on top and the 27 on bottom .

When i work it out I get this...

= 1 1
____ * ____
-27m^6 n^3

Then I dont get why the m and n are on top and 27 is on bottom. Unless the n cross multipleis up to the left and the m^6 only cross multiplies to the other side... But thats probably not it.

9. Aug 2, 2011

### Nelo

So.. you solved it like multiplication then did a recipriocol of the "9" for the final step..?

10. Aug 2, 2011

### gb7nash

[tex ]\frac{1}{-27m^6}*\frac{1}{n^3}[/tex ]

(without the space in the tex tags) You will save yourself and anyone who's reading your work lots of time.

11. Aug 2, 2011

### Nelo

$$\frac{1}{-27m^6}*\frac{1}{n^3}$$

12. Aug 2, 2011

### eumyang

But if you "distribute" the -3 exponent outside the parentheses first, you would get
$\left(-3m^{-3}n^{-1}\right)^{-3} = (-3)^{-3}m^{9}n^{6}$
It's much easier to simplify from this, as opposed to using the reciprocal method to start.

13. Aug 2, 2011

### Nelo

Reopen my tinypic link, and look at 3k) . How do i solve that using your method? I simplified the top half then tried solving it and it failed.

I get to -10s^5 t^3
__________
4s^2 t^3

.

I get it this far. Now what am i supposed to do? ( I know im not using your tex thing.. I dont know how and i dont really have time at the moment)

Last edited: Aug 2, 2011
14. Aug 2, 2011

### Nelo

Ok, well whatever I guess. Just answer me this then. If there is an expression that include both negetive and positive exponents, I keep the positive exponents where they are and recipricol the negitive ones yes?

15. Aug 2, 2011

### eumyang

This is wrong. It's supposed to be:
$$\frac{-10s^{-5}t^3}{4s^2 t^{-3}}$$
Now you can either subtract the exponents, or "flip" the variables that contain the negative exponents across the bar and make the exponents positive, like this:
$$\frac{-10 t^3 t^3}{4s^2 s^5}$$
I'll leave you to do the rest.

And if you must know, I didn't reply right away because it's morning here and I have to get ready for the day. The rest of us aren't chained to our computers 24 hours a day, you know. Now I have to run. Good luck.