Solving Equations with exponents

1. Sep 23, 2004

preet

Hi... I'm new...
I was having problem with these questions:
(a^3 x b)^2 (-a/b)^3
I ended up with a^-9(b)... is that wrong or right?
and
(x/-y)^3 (-xy)^4
I got up to (-x^7 y^4 )/ -y^3
I don't know what to do here. Thanks!

2. Sep 23, 2004

christinono

For the first one, you did not do it right. Remember to keep the negative sign. Also, when you multiply like bases with exponents, you add the exponents; you don't multiply them

3. Sep 23, 2004

christinono

As for the second one, you started out just fine. All that's left now is to simplify it. Remember, when you divide like bases, you SUBTRACT the exponents. So the exponent on y would be 4-3=1.

4. Sep 23, 2004

preet

still having trouble understanding this...
(a^3*b)^2
=a^6 * b^2

(-a / b) ^3
= -a^3 / b^3

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

Not sure what I'm doing from here... confused with
a^6 * -a^3 ... are they considered like terms? can I add them (exps)?
b^2 / b^3 = b^-1

For the second one, what happens to the minus sign on -y^3? ... the final answer should be -x^7 * y ... so yeah, what happened to that minus sign?

Last edited: Sep 23, 2004
5. Sep 23, 2004

christinono

Sorry, I made a mistake in the first one when I did it (stupid...)
OK, now that you have a^6 * b^2 * -a^3 / b^3, bring the negative to the front of the expression to make it simpler to look at. Now you can combine the a's and the b's, adding their exponents. Do you get it?

6. Sep 23, 2004

preet

What do you mean by in front of the equation?...

7. Sep 23, 2004

christinono

sorry, I didn't mean to confuse you.
I mean write it out like this: -a^6 * b^2 * a^3 / b^3
Now for the a's: the sum of their exponents is 6+3
for the b's: the sum of their exponents is 2+(-3)
(remember that when a base is in the denominator, it's exponent needs to be multiplied by -1 if it is to be placed in the numerator. If you are totally confused by my last sentence, you can just add the exponents of the b's this way: sum = 2-3 (since you are dividing, you subtract the exponents)

Make more sense?

8. Sep 23, 2004

preet

Still kind of iffy with the moving of the negative sign... that's what's confusing me in both questions. How can you move the negative sign in the first question to a^6? ... that would mean 'a' becomes positive because the exponent is an even number right?

If you do -a^6 * a^3 I understand what the resulting exponent will be but I don't know about the term ... what happens to -a and a...

looking over what I wrote in the first post, it was a typo... my answer was
a^9 * b... b goes on top of the fraction because it was to the power of -1... but what happened to a? I know that the rule is x^n * x^m = x^n+m, but what do I do in my case (where the 'a' is negative)?

This same problem is happening in the second question where I divide y^4 by -y^3... I don't know what to do. Thanks for the help so far. I really apprecieate it =)

9. Sep 23, 2004

Tom McCurdy

Christinono first of all welcome to PF second of all I might suggest learning how to you Latex typesetting on the forum to make the math look cleaner
you type [*tex] ax^b [*/tex] with out * and you will get $$ax^b$$

10. Sep 23, 2004

Tom McCurdy

Example of latex
$$-a^6* b^2*(a^3/b^3)$$

11. Sep 23, 2004

christinono

Thanks Tom!
I just joined the forum a few days ago and have not had time to learn the
Latex code yet. In a few days, I should have it down...

12. Sep 23, 2004

Tom McCurdy

13. Sep 23, 2004

preet

I've been to that site before... what I'm looking for isn't there (or I can't find it)

14. Sep 23, 2004

christinono

Let me try to explain this to you:
When you have a bunch of bases with exponents that are multiplied or divided, you can place the negative wherever without changing the value of the expression.
eg: (-2)(3)(9)= -54
now, if you put the negative somewhere else,
(2)(-3)(9) , it still gives -54.
The same goes for bases and exponents.
When you multiply -a^2 by a^3 (just an example), you add the exponents (it gives you 5), then you decide if the expression will be negative or positive. A negative times a positive gives a negative, so your answer is -a^5.
As long as the bases are the same letter or number, you can multiply them by adding their exponents, even though one is negative.

15. Sep 23, 2004

preet

Then why does this: $$(x/-y)^3 (-xy)^4$$
end up becoming $$-x^7 * y$$?

1) $$x^3 / -y^3 * -x^4 * y^4$$

2) $$x^3 * -x^4 = -x^7$$ ... $$y^4 / -y^3 = -y$$

3) So now what? $$-x^7$$... then where does the -y go and how does it become y?

I'm just lost... plain and simple. And I still don't know if the first one is right or wrong.

16. Sep 23, 2004

Tom McCurdy

I was almost trickerd into $$x^7 * y$$ but let me show how i got $$-x^7 * y$$

$$(x/-y)^3 = (-x/y)^3$$
=
$$(-x^3/y^3)*x^4y^4$$ Since even powers get rid of negitive sign
=
$$-x^(3+4)*y(4-3)$$
=$$-x^7 * y$$

Last edited: Sep 23, 2004
17. Sep 23, 2004

preet

Thanks! I just wanted to confirm, does that mean that $$x^7 * -y$$
is also correct? (because the answer will end up neg. if y is > 0 in both cases).

18. Sep 23, 2004