I am a bit confused about the exponent rules... I just need to simplify these problems.

ab²∗a³b=

(ab)⁴=

x⁵∕y⁴∗(y∕x)³=

Thank you so much for your help!!!

~Winter

Winter's Twelfth Night says...

I am a bit confused about the exponent rules... I just need to simplify these problems.

ab²∗a³b=

(ab)⁴=

x⁵∕y⁴∗(y∕x)³=

Thank you so much for your help!!!

~Winter

ab²∗a³b=

(ab)⁴=

x⁵∕y⁴∗(y∕x)³=

Thank you so much for your help!!!

~Winter

Mamillius: Merry or sad shall’t be?

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Rosey Unicorn says...

When multiplying exponents, just add them. When dividing them, subtract. When your multiplying the group [the (ab)4 problem] it's just a case of figuring out how many a's need to be multiplied. So, in that problem, it's four.

Get it?

Get it?

You know you're a writer when you're not alarmed at hearing voices in your head, you can't read a book without analyzing it for plot & characters and you consider something you nearly killed yourself to write the most rewarding.

Guilty as charged.

Guilty as charged.

Winter's Twelfth Night says...

Thank you so much for your help! I'm still a little confused about multiplying the four a's. Would the answer just be ab⁴? Thanks again for replying!

~WInter

~WInter

Mamillius: Merry or sad shall’t be?

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Rosey Unicorn says...

Close. It'd be a4b4 (Pretend those are subscripts, my computer won't let me do it) Since your multiplying both the a and the b.

You know you're a writer when you're not alarmed at hearing voices in your head, you can't read a book without analyzing it for plot & characters and you consider something you nearly killed yourself to write the most rewarding.

Guilty as charged.

Guilty as charged.

Winter's Twelfth Night says...

Oh ok! I get it now. Thank you sooooo much for the help!

Mamillius: Merry or sad shall’t be?

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Rosey Unicorn says...

Your welcome!

This stuff took me forever to learn too.

This stuff took me forever to learn too.

You know you're a writer when you're not alarmed at hearing voices in your head, you can't read a book without analyzing it for plot & characters and you consider something you nearly killed yourself to write the most rewarding.

Guilty as charged.

Guilty as charged.

Winter's Twelfth Night says...

Yes, I definitely don't have a math mind.

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

Incandescence says...

Winter:

I'll go ahead and carry the first problem out in full, if you don't mind.

ab²∗a³b=a¹∙b²∙a³∙b¹

By the associative property (i.e., abc=(ab)c=a(bc)), we can rewrite this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹

Since associative multiplication is commutative (i.e., ab=ba), we can write this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹

and by associativity we can say:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)

Since the rule of exponent multiplication is addition, we get:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)=(a¹⁺³)(b²⁺¹)=a⁴b³

The second problem is identical to the first problem: just make sure you understand that everything inside the parentheses is put to the exponent. That is, if I had (ab)³, I would say, (ab)³=(ab)(ab)(ab), by definition of exponentiation. From there, I would follow the procedure I gave you above, changing the parentheses around and use commutativity to get my result.

Best,

Brad

I'll go ahead and carry the first problem out in full, if you don't mind.

ab²∗a³b=a¹∙b²∙a³∙b¹

By the associative property (i.e., abc=(ab)c=a(bc)), we can rewrite this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹

Since associative multiplication is commutative (i.e., ab=ba), we can write this as:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹

and by associativity we can say:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)

Since the rule of exponent multiplication is addition, we get:

ab²∗a³b=a¹∙(b²∙a³)∙b¹=a¹∙(a³∙b²)∙b¹=(a¹∙a³)∙(b²∙b¹)=(a¹⁺³)(b²⁺¹)=a⁴b³

The second problem is identical to the first problem: just make sure you understand that everything inside the parentheses is put to the exponent. That is, if I had (ab)³, I would say, (ab)³=(ab)(ab)(ab), by definition of exponentiation. From there, I would follow the procedure I gave you above, changing the parentheses around and use commutativity to get my result.

Best,

Brad

"If I have not seen as far as others, it is because giants were standing on my shoulders." -Hal Abelson

Winter's Twelfth Night says...

Thank you for writing it out! That makes a lot of sense. So you just write it out as a longer problem so that you can add up the like bases. That's what you were saying, right? At least that's the way that makes the most sense to me. Anyway, thanks again!

-Winter

-Winter

Hermione: As merry as you will.

Mamillius: A sad tale’s best for winter. I have one

Of sprites and goblins.

The Winter's Tale

thunder_dude7 says...

It seems that you've been helped, but I'm bored.

ab²∗a³b=

Start by looking at the variables individually. Let's start with "a". a^3 multiplied by a, written out, is...

a x a x a x a

That equals a^4.

The same with the Bs means that they're equal to b^3.

(ab)⁴=

Well, this can be written as "(ab) x (ab)". This is a double distributive property thing. First, distribute the a. That makes the second part a^2 x ab. Next, distribute the b. That makes it a^2b x ab^2.

Well, I have to go. Can't do the third one. Sorry.

ab²∗a³b=

Start by looking at the variables individually. Let's start with "a". a^3 multiplied by a, written out, is...

a x a x a x a

That equals a^4.

The same with the Bs means that they're equal to b^3.

(ab)⁴=

Well, this can be written as "(ab) x (ab)". This is a double distributive property thing. First, distribute the a. That makes the second part a^2 x ab. Next, distribute the b. That makes it a^2b x ab^2.

Well, I have to go. Can't do the third one. Sorry.

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