1. Feb 6, 2006

### runicle

I have the equation [(m^3)(n^-3)]^-1 over m^-5n
So far i got 1 over m^6n^6 divide by m^-5n...
Is it right or no?

2. Feb 6, 2006

### Tom Mattson

Staff Emeritus
I've moved this from the Tutorials Forum into the Homework Help section.

That's not an equation, it's an expression.

No, it isn't. What steps did you take to manipulate the numerator of your original expression?

3. Feb 6, 2006

### runicle

First off i went to the basics I multiplied (m^3)(n^-2) and got m^6n^6
Then i inversed it to get 1 over m^6n^6, so i can divide that with m^-5. Yet, i still don't know arithmetic.

4. Feb 6, 2006

### Tom Mattson

Staff Emeritus
That's not right. You should be using the following rule:

$$(x^ay^b)^c=x^{ac}y^{bc},$$

which should be in your textbook.

5. Feb 6, 2006

### chroot

Staff Emeritus
What are you trying to do? Simplify the expression?

Note that $m^3 \cdot n^{-3}$ does not equal $m^6 n^6$.

Where did the extra powers of m come from? Where did the negative powers of n go?

- Warren

6. Feb 6, 2006

### runicle

So the exponent doesn't affect the base number

7. Feb 6, 2006

### chroot

Staff Emeritus
No. There is no "base number" here anyway -- just the variables m and n.

- Warren

8. Feb 6, 2006

### runicle

Im so confused

9. Feb 6, 2006

### runicle

Okay wait what happens to the exponents when its (m^3)(n^2)

10. Feb 6, 2006

### chroot

Staff Emeritus
Did you see what Hurkyl posted? The general rule of "exponent distribution:"

$$(x^ay^b)^c=x^{ac}y^{bc}$$

You have an example of this here:

$$(m^3 n^{-3})^{-1} = m^{3 \cdot -1}n^{-3 \cdot -1} = m^{-3} n^{3}$$

Again, I'll ask you: what are you trying to do? Simplify the expression?

- Warren

11. Feb 6, 2006

### chroot

Staff Emeritus
Nothing at all "happens" to the exponents -- the bases are different, and thus they are completely unrelated to each other.

- Warren

12. Feb 6, 2006

### runicle

Would it be mn^-6?

13. Feb 6, 2006

### runicle

Sorry i just read simplify. yes i need to simplify

14. Feb 6, 2006

### chroot

Staff Emeritus
No. Now you're getting me a bit confused. First you told us the problem was this:

(m^3)(n^-3)]^-1 over m^-5n

and now you're talking about (m^3)(n^2). Which is it?

Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

- Warren

15. Feb 6, 2006

### chroot

Staff Emeritus
No. Now you're getting me a bit confused. First you told us the problem was this:

(m^3)(n^-3)]^-1 over m^-5n

and now you're talking about (m^3)(n^2). Which is it?

Please repeat the problem, taking care to type it exactly as shown in your homework. Also, please tell us exactly what you're trying to do: are you just supposed to simplify the expression?

- Warren

16. Feb 6, 2006

### runicle

Nevermind i got it I'm simplifying it It's n^2m^5 over m^3n

The question was simplify [(m^3)(n^-3)]^-1 over m^-5n

17. Feb 6, 2006

### runicle

I got another question I have to factor 3x^2 -13x-10
So far i get this far
3x^2 -13x-10
3x^2+2x-15x-10
x(3x+2)-5(3x+2)
What do i do next...

18. Feb 6, 2006

### chroot

Staff Emeritus
I'm afraid that

$\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}$

does not simplify to

$\frac{n^2 m^5}{m^{3n}}$

- Warren

19. Feb 6, 2006

### runicle

Well i don't know any other way

20. Feb 6, 2006

### chroot

Staff Emeritus
Let's do this in steps.

First, simplify the numerator of

$\frac{(m^3 n^{-3})^{-1}}{m^{-5n}}$

The -1 exponent "distributes", multiplying the exponents on m and n:

$(m^3 n^{-3})^{-1} = m^{-3}n^3}$

So the entire expression becomes:

$\frac{m^{-3}n^3}{m^{-5n}}$

Now, factors with negative exponents can be "flipped" across the division line, as can be seen by multiplying both sides by the same factor with the exponent positive rather than negative:

$\frac{m^{-3}n^3}{m^{-5n}} = m^{-3}n^3 m^{5n}$

One negative-exponent factor went up, while the other negative-exponent factor I left alone.

Finally, multiplyiing two factors (with the same base) is the same as adding their exponents:

$n^3 m^{5n - 3}$

If you do not understand any step of this, please let me know.

- Warren

Last edited: Feb 6, 2006