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(3x^3)^2(2x)^3 - QuestionCove
OpenStudy (anonymous):

(3x^3)^2(2x)^3

4 years ago
OpenStudy (anonymous):

Here's what I've got so far

4 years ago
OpenStudy (anonymous):

\[(3x ^{3})^{2} (2x)^{3} \] 3*2 gives me \[3x^{6} (2x) ^{3}\]

4 years ago
OpenStudy (anonymous):

If I times 3&2 that gives me 6 Adding 6&3=9 So that gives me 6x^{9}

4 years ago
OpenStudy (uri):

Rightt :)

4 years ago
OpenStudy (anonymous):

But When I checked it to see if i was right (i subsittuted something for the x It didn't add up right

4 years ago
OpenStudy (anonymous):

@uri Come backkkkkkk!

4 years ago
OpenStudy (anonymous):

Great your back :D

4 years ago
OpenStudy (uri):

Let me see

4 years ago
OpenStudy (uri):

Did you put 6x^9?

4 years ago
OpenStudy (anonymous):

Yep

4 years ago
OpenStudy (uri):

Try putitng 6x^18

4 years ago
OpenStudy (anonymous):

Still comes up different

4 years ago
OpenStudy (uri):

@terenzreignz

4 years ago
OpenStudy (terenzreignz):

What are we trying to do here?

4 years ago
OpenStudy (anonymous):

Call in reinforcements! :P

4 years ago
OpenStudy (anonymous):

Simplifying the original expression

4 years ago
OpenStudy (terenzreignz):

Laws of Exponents

4 years ago
OpenStudy (terenzreignz):

\[\LARGE (3x^3)^2(2x)^3\]

4 years ago
OpenStudy (anonymous):

Hey how did you make that bigger? I was trying that but I couldn't make it readable!

4 years ago
OpenStudy (anonymous):

off-topic sorry

4 years ago
OpenStudy (terenzreignz):

\large \Large \LARGE \huge \Huge ^ varying degrees of size

4 years ago
OpenStudy (anonymous):

thanks :D

4 years ago
OpenStudy (anonymous):

anyways...

4 years ago
OpenStudy (anonymous):

Back to the problem

4 years ago
OpenStudy (terenzreignz):

<nods> Yes, the laws of exponents... Much like how multiplication distributes over addition, exponentiation distributes over multiplication. Let me show you what I mean: \[\Large (ab)^m\] You just raise all the factors to the exponent Like this \[\Large a^mb^m\]

4 years ago
OpenStudy (terenzreignz):

With that in mind, you can now easily simplify this bit: \[\LARGE (3x^3)^2\color{red}{(2x)^3}\]

4 years ago
OpenStudy (anonymous):

So that equals \LARGE 2 ^{3} * x^{3}

4 years ago
OpenStudy (anonymous):

sorry about the formatting 2^3 *x^3

4 years ago
OpenStudy (terenzreignz):

Yeah... you should enclose it in \[\Large \text{\[ <stuff>\]}\]

4 years ago
OpenStudy (anonymous):

ah thanks

4 years ago
OpenStudy (terenzreignz):

And yes, you are correct, now \[\Large 2^3 = \color{red}?\]

4 years ago
OpenStudy (anonymous):

Hey math lesson and latex!

4 years ago
OpenStudy (anonymous):

That would be 8

4 years ago
OpenStudy (terenzreignz):

That's right. So that bit: \[\LARGE (3x^3)^2\color{red}{(2x)^3}\]just becomes \[\LARGE (3x^3)^2(\color{red}{8x^3})\] correct?

4 years ago
OpenStudy (anonymous):

Alrighty! That makes sense

4 years ago
OpenStudy (terenzreignz):

Now, do the same to this part: \[\LARGE \color{blue}{(3x^3)^2}(\color{}{8x^3})\]

4 years ago
OpenStudy (anonymous):

OK So do I add the 3&2 first?

4 years ago
OpenStudy (terenzreignz):

Consider \(\Large x^3\) as a single entity for now. Just tell me what happens to that 2-exponent.

4 years ago
OpenStudy (anonymous):

The 2 is added after \[3x^{3}]/ is solved

4 years ago
OpenStudy (terenzreignz):

No, I mean, what happens here \[\LARGE \color{blue}{(3x^3)^2}(\color{}{8x^3})\] as per this rule: \[\Large (ab)^m = a^mb^m\]

4 years ago
OpenStudy (terenzreignz):

and it's \] at the end.

4 years ago
OpenStudy (anonymous):

OH! so that means 3^3*x^3

4 years ago
OpenStudy (terenzreignz):

Why does 3 get cubed? Was the exponent 3?

4 years ago
OpenStudy (anonymous):

27^3?

4 years ago
OpenStudy (anonymous):

And then I times 2*3

4 years ago
OpenStudy (terenzreignz):

Time out

4 years ago
OpenStudy (anonymous):

So that gives me 27^6

4 years ago
OpenStudy (anonymous):

27^6 *8^3

4 years ago
OpenStudy (terenzreignz):

Let's study this a little more closely, because your answers are (frankly) getting more and more outrageous... \[\Large (3x^3)^2\]

4 years ago
OpenStudy (anonymous):

Err sorry about that, this is a new topic for me

4 years ago
OpenStudy (terenzreignz):

Understood, but this is one of the more basic, so I really don't want you to be having a hard time with this... because your speed at your more intermediate maths depends on whether or not this stuff is just second nature to you...so...

4 years ago
OpenStudy (anonymous):

I understand!

4 years ago
OpenStudy (terenzreignz):

Okay, so let's study this closely: \[\Large (3x^3)^2\] Now, I noticed you had no difficulty whatsoever in simplifying that other one \[\Large (2x)^3=2^3x^3=8x^3 \]

4 years ago
OpenStudy (anonymous):

Right the 3 is distributive inside the parenthesis. Now that we have a exponent inside & outside the parenthesis, which do i solve first?

4 years ago
OpenStudy (terenzreignz):

I'm glad you can identify your problem :) Maybe, to get you started, we do a simple substitution.. just so you know, in a manner of speaking, where your priorities lie. Let's let \[\Large u = x^3\] so that \[\Large (3x^3)^2=(3u)^2\] okay?

4 years ago
OpenStudy (anonymous):

OK! So I have to solve the exponint inside the parenthisis first! next to the varible the expoint is closer to!

4 years ago
OpenStudy (terenzreignz):

Or, here's what I think would be the easier way to look at things... Let's look at that example you handled so well:

4 years ago
OpenStudy (anonymous):

OK

4 years ago
OpenStudy (terenzreignz):

\[\Large (2x)^3\]

4 years ago
OpenStudy (anonymous):

Right I distributed the 3 inside the parentheses

4 years ago
OpenStudy (terenzreignz):

What the exponent outside actually does is MULTIPLY ITSELF to all the exponents inside the parenthesis. Now, it's understood that when there's no exponent, that the exponent is 1...so...\[\Large (2^1x^1)^3\]

4 years ago
OpenStudy (anonymous):

OK! So that's where the x^2 came from!

4 years ago
OpenStudy (terenzreignz):

So, 3 multiplies itself to all the EXPONENTS: \[\Large 2^{1\times 3}x^{1\times3}= 2^3x^3 = 8x^3\]

4 years ago
OpenStudy (terenzreignz):

Now, you can do likewise with \[\LARGE (3x^3)^2\]Only this time, not all the exponents are 1. No big deal, the same principle applies... go ahead now...

4 years ago
OpenStudy (terenzreignz):

Just remember: The exponent *outside* the parentheses multiplies itself to each of the exponents *inside* the parentheses...

4 years ago
OpenStudy (anonymous):

OK So that means \[3^{1*2} & x ^{3*2}\]

4 years ago
OpenStudy (anonymous):

Dang it

4 years ago
OpenStudy (terenzreignz):

Never mind, it seems you got the correct idea :)

4 years ago
OpenStudy (anonymous):

So that gives me

4 years ago
OpenStudy (terenzreignz):

\[\Large 3^{1\times 2}x^{3\times 2}\] simplify?

4 years ago
OpenStudy (anonymous):

Thanks

4 years ago
OpenStudy (anonymous):

OK so that gives me 3^2 * x^6

4 years ago
OpenStudy (anonymous):

So that would give me 3x^8?

4 years ago
OpenStudy (terenzreignz):

no... hold it

4 years ago
OpenStudy (terenzreignz):

Correct up to this point: \[\Large 3^2x^6\] Now how on earth does this become \[\Large 3x^8\]??

4 years ago
OpenStudy (anonymous):

OH! I'm adding instead of multiplying Should it become 3x^12?

4 years ago
OpenStudy (terenzreignz):

No.

4 years ago
OpenStudy (anonymous):

Hold on let me check the other equation

4 years ago
OpenStudy (terenzreignz):

You don't move exponents around like that unless they have the same base!!!!

4 years ago
OpenStudy (terenzreignz):

3 and x? NOT the same base.

4 years ago
OpenStudy (anonymous):

Oh never mind

4 years ago
OpenStudy (terenzreignz):

Anyway, if we went by that logic, we should get \[\Large 2^3x^3 = 2x^6 \] or something... clearly wrong

4 years ago
OpenStudy (anonymous):

Right that's what I did

4 years ago
OpenStudy (terenzreignz):

How did we simplify \[\Large 2^3x^3\]again?

4 years ago
OpenStudy (anonymous):

We just added the two together and used the same exponint

4 years ago
OpenStudy (anonymous):

So we got 2^3

4 years ago
OpenStudy (anonymous):

2x^3

4 years ago
OpenStudy (terenzreignz):

2^3 which is?

4 years ago
OpenStudy (anonymous):

8

4 years ago
OpenStudy (anonymous):

OH! OK! So 3^2 * x^6

4 years ago
OpenStudy (terenzreignz):

Because 2^3 is jut a number.

4 years ago
OpenStudy (terenzreignz):

And so is 3^2.

4 years ago
OpenStudy (anonymous):

9x^6

4 years ago
OpenStudy (terenzreignz):

yes. (Really, before you should even think about doing algebra, make sure your grounding in Arithmetic is solid)

4 years ago
OpenStudy (terenzreignz):

So ultimately, we get \[\LARGE \color{blue}{(3x^3)^2}(\color{}{8x^3})=(\color{blue}{9x^6})(8x^3)\]

4 years ago
OpenStudy (anonymous):

Yes I suppose I should have payed more attention in pre algebra

4 years ago
OpenStudy (anonymous):

Thank you!

4 years ago
OpenStudy (anonymous):

*Longest openstudy question for me to date*

4 years ago
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