Giving Lessons: Multiplying Exponents

Question

aihberkhan · Answer

If you don’t already know… for the past few days I have decided to start Giving Lessons. If you want to see the rest of them just go on my profile, click on my questions, and click on the ones that start with: “Giving Lessons”. I feel that this is definitely helping some of you, so I decided that this is a great way to help you guys out! If you already know this, then tag someone who doesn’t, but if you don’t already know this, then hopefully this will help you out!

Let’s Get Started!

The first sample problem we are going to be working with is

$$2m^2 * 2m^3$$

This is an easy problem to start with so I think it would be good if you are a beginner. 

First we should multiply the two coefficients to get that out of the way. So we would do:

$$2 * 2$$

When completing that we should get 4 so we know that our answer will start with a 4.

Now when multiplying exponents all you do is add them! It’s that easy! 

So our exponents are 2 and 3. When adding 2 and 3 you should get 5! That’s it! We now can get our answer. So when putting all those calculations together… our answer should be:

$$4m^5$$

DONE! GREAT JOB! :)

—————————

Still Confused? Don’t worry! We will be doing another sample problem!

—————————

Okay, our next sample problem is:

$$m^4 * 2m ^{-3}$$

All we do is add 4 and -3!

When we do this we should get 1! Whenever you have an exponent of 1, you don’t actually have to write the 1. 

So you wouldn’t write it like this:

$$m^1$$

Instead you would just write:

$$m$$

Now our complete answer would be:

$$2m$$

DONE! GREAT JOB! :)

—————————

Still Confused? Don’t worry! We will be doing one last sample problem!

—————————

Our last sample problem is:

$$2x ^{3}y ^{-3} * 2x ^{-1}y ^{3}$$

This is a bit longer than the others, but it is as easy as the other ones. All we did in this sample problem is add another variable, which is “y”. 

Now lets focus on the exponents on the two variables x. 

We have a 3 and a -1. 

When adding together 3 and -1 we should get 2. So we know the first part of our answer. 

Now when looking at the exponents on the two variables y we have a -3 and 3 . 

When adding together 3 and -3 we get 0! That means that the y cancels out of this problem. So, we shouldn’t see the variable y anymore!

Lastly, our coefficients are 2 and 2. So we need to multiply those. When doing so we should get 4!

Now that we have all our calculations we can finally put them together and get our answer!

We should get:

$$4x^2$$

DONE! GREAT JOB! :) 

—————————

Still Confused? Don’t Worry! If you have any questions just comment them down below! I will try my best to answer them as soon as possible! 

—————————

I hope this helped you in some way! Once again, If you already knew this, then tag someone who doesn’t. But if you didn’t and this helped you… then I am very glad that I helped you out! 

Alright, bye now! :)

denonakavro · Answer

Is this all yours ?

anonymous · Answer

We can even expand this from more basic lessons.

$x \times x \times x = x^3 = 1\times x^1 \times 1 \times x^1 \times 1 \times x^1 $ 
this gives the additive property of exponents when multiplying bases either variable or coefficient or both.

anonymous · Answer

properties of multiplication is that they yield the same result no matter what part of expression is multiplied first. 

$2 \times 3 \times 5 = 5 \times 2 \times 3 = 3 \times 5 \times 2  $

aihberkhan · Answer

Yes it is! :) 

@denonakavro

aihberkhan · Answer

Great! :)

@panlac01