Solving Equations with exponents

[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!

 

[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

 

[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.

 

[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?

 

[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?

 

[preet]

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

 

[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?

 

[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 =)

 

[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] without * and you will get $$ax^b$$

 

[Tom McCurdy]

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?

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

 

[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...

 

[Tom McCurdy]

Here is a good site to start you out with learning exponets preet
http://www.purplemath.com/modules/exponent.htm

 

[preet]

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

 

[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.

 

[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.

 

[Tom McCurdy]

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

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
=
$$-x^(3+4)*y(4-3)$$
=$$-x^7 * y$$

 

[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).

 

[recon]

Yes, your answer is correct, preet.