MHB Applying the rules for exponents

[MagicalGoose]

Hi Guys and Grillz
Im not sure how these forums work, but if someone could help me real quick by explaining me how i should count this (3x2y)^8 (2xy)^3. I am sorry if i asked a dumb question or posted it in the wrong place. This is my first time doing this. :D

 

[Greg]

First, simplify inside the brackets:$$(3x2y)^8(2xy)^3=(6xy)^8(2xy)^3$$Then distribute the exponent over the terms in brackets:$$(6xy)^8(2xy)^3=6^8x^8y^8\cdot2^3x^3y^3$$Evaluate the constant terms (the ones that aren't variables):$$6^8x^8y^8\cdot2^3x^3y^3=1679616x^8y^8\cdot8x^3y^3=13436928x^8y^8x^3y^3$$Now simplify the variables by adding the exponents of like terms:$$13436928x^8y^8x^3y^3=13436928x^{11}y^{11}$$. . . and that's it!

 

[MagicalGoose]

First, simplify inside the brackets:$$(3x2y)^8(2xy)^3=(6xy)^8(2xy)^3$$Then distribute the exponent over the terms in brackets:$$(6xy)^8(2xy)^3=6^8x^8y^8\cdot2^3x^3y^3$$Evaluate the constant terms (the ones that aren't variables):$$6^8x^8y^8\cdot2^3x^3y^3=1679616x^8y^8\cdot8x^3y^3=13436928x^8y^8x^3y^3$$Now simplify the variables by adding the exponents of like terms:$$13436928x^8y^8x^3y^3=13436928x^{11}y^{11}$$. . . and that's it!

Thanks but i actually typed it wrong XD I should have been like this : \left({3x}^{2}y\right)^{8}\left({2}^{x}y\right)5
I wasn't sure how this site works at first so yeah, allot of bad typping was involved.

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I think i got it rigth this time.

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(3x^2y)^8(2xy)^5 OK. This is too complicated for me to type it with the sites symbols so i'll just do it like this.

 

[Greg]

If you post your work we can help you identify and correct any errors that may be present.$$\text{ }$$You need to wrap your latex code in $$$$ tags.

 

[MagicalGoose]

I hope this works.

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$$\left({3x}^{2} y\right){}^{8}\left(2xy \right){}^{5} $$ Ok this should be the right way to write it :D Thank you for telling me how to do it.

 

[MagicalGoose]

$$\left({3x}^{2} y\right){}^{8}\left(2xy \right){}^{5} $$ Guys can anyone of you tell me how i should be doing this one step by step?

 

[MarkFL]

There are three rules for exponents you want to apply here:

[box=blue]$$(ab)^c=a^cb^c\tag{1}$$

$$\left(a^b\right)^c=a^{bc}\tag{2}$$

$$a^b\cdot a^c=a^{b+c}\tag{3}$$[/box]

Can you proceed?

 

[MarkFL]

$$\left({3x}^{2} y\right){}^{8}\left(2xy \right){}^{5} $$ Guys can anyone of you tell me how i should be doing this one step by step?

I didn't notice until after posting help that you posted the same question in more than one thread. We ask that you post a question only once because not only is this redundant, it can lead to duplication of effort which wastes the time of our helpers.

We also ask that threads be given a title that briefly describes the question being asked. :D

 

[MagicalGoose]

There are three rules for exponents you want to apply here:

[box=blue]$$(ab)^c=a^cb^c\tag{1}$$

$$\left(a^b\right)^c=a^{bc}\tag{2}$$

$$a^b\cdot a^c=a^{b+c}\tag{3}$$[/box]

Can you proceed?

I tried it but i don't know if i did it right.

 

[MarkFL]

I tried it but i don't know if i did it right.

Well, since you haven't posted your work, I don't know if you did it correctly either. :D

 

[MagicalGoose]

$$ 209952 {x}^{21} {y}^{13} $$ This is what i did so far.

 

[MarkFL]

$$ 209952 {x}^{21} {y}^{13} $$ This is what i did so far.

Yes, that's correct. (Yes)

 

[GeraldArgue]

To solve the problem: (3x2y)^8 (2xy)^3, we have to use the formula i.e.
a^b.a^c=a^b+c.

=6xy^8.2xy^3
=12xy^8+3
=12xy^11
=12^11.x^11.y^11
=13436928x^11y^11
The above is the solution for the question.

 

[MarkFL]

To solve the problem: (3x2y)^8 (2xy)^3, we have to use the formula i.e.
a^b.a^c=a^b+c.

=6xy^8.2xy^3
=12xy^8+3
=12xy^11
=12^11.x^11.y^11
=13436928x^11y^11
The above is the solution for the question.

Your first line should read:

=(6xy)^8(2xy)^3

However, please do not simply post in threads that have already been answered fully to rehash what has already been posted. If you have a different method to post or new insights to share, then that's great, but to simply post things that have already been posted adds no value to a thread. :D