# Lesson video

In progress...

Hello, my name is Mr Clasper, And today we're going to look at the division law for indices.

Let's try this example, three to the power of six divided by three to the power of four.

We could rewrite this as a fraction, so three to the power of six, divided by three to the power of four, and this could also be written like this.

Now what we can think about, every time we divide three by three in this fraction, it will give us a value of one and therefore it will cancel out.

So if we divide three by three, that will equal one.

We divide another three by three, this will equal one.

Dividing another three by three will equal one, and dividing another three by three would give us another one.

That leaves us with a value of three to the power of two.

What happened to the base, and what happened to the indices? Well we can see that the base stayed the same and the indices were subtracted.

Let's have a look at this example, we have X to the power of six this time, divided by X to the power of four.

I could write this as a fraction.

And what we can think about is every time I divide X by X, or anything by itself, it will give a value of one.

So X divided by X would give us one and cancel out, X divided by X will give us one again, so this would cancel out.

X divided by X would give us a value of one and another X divided by X will give us a value of one.

This leaves us with X to the power of two.

What happened to the base, and what happened to the indices? Again, looking closely, we can see that the base remained the same, and the indices were subtracted, so six subtract four gave us a value of two.

This is our general algebraic rule.

So if we have X to the power of A, divided by X to the power of B.

What happens to the base? Well we know that the base remains the same.

And what happens to the indices? From the previous examples, we know that we subtract the indices.

So our final response would be X to the power of A minus B.

And this is our general rule.

Here are some questions for you to try.

Pause the video to complete your task, and click resume once you're finished.

So just be careful with a couple of these, so question one D, we can see we have D to the power of seven, divided by D to the power of seven, so that means we'd have to subtract these powers, therefore, seven subtract seven will give you a power of zero.

And looking to question two, we can see that the first example is false, so the correct answer is actually K to the power of four.

So eight, subtract four, would give us a new power of four.

Also, the example underneath this, is also false, so this should be the other way around.

So we should have a multiplication involving five K's as a numerator, and a multiplication using three K's as a denominator.

Let's try this example.

Nine X to the power of six divided by three X to the power of four.

We could write this as a fraction.

And thinking about cancelling factors from the numerator and the denominator, these will all give us a value of one when we divide them by one another.

The last thing we need to consider, is calculating nine divided by three.

This will give us a value of three.

So our final answer should be three multiply by X, multiply by X or, three X squared.

What happened to the base? Well once again, the base stayed the same.

What happened to the indices? The indices were subtracted again, so six subtract four, gave us our power of two.

What happened to the coefficients? The coefficients were divided.

So we divided nine by three, to get our new coefficient of three.

Let's have a look at the general algebraic rule.

So what happens to the base? At the moment, we have a base of X.

And the base and our solution will be the same.

So our response, would have X as a base.

What happens to the indices? We would have to subtract the indices as we did before.

And what happens to the coefficients? We must divide them.

So we would have to divide P by Q.

That means our final response for this general rule, would be P over Q X to power of A minus B.

Here's some questions for you to try, pause the video to complete your task and click resume once you're finished.

So let's have a look at a couple of these, so we have three C, we have an answer of three H to the power of two, so remember we're going to divide our coefficients, and then subtract our powers.

Likewise for F, this is an interesting one.

We got 19 V to the power of five, divided by 19 V to the power of four, so we divide our coefficients again, which would give us a value of one, and we subtract our powers, which would also give us a power of one.

So therefore, our final answer is V to the power of one or V.

If you really want to you could write one V as well, there's no harm, but it's not necessary.

And if you look at question four, we can see our two complete answers in the middle, so the box on the left, would simplify to one third, of A to the power of six, so that's not what we want.

And the box at the end, would simplify to three over two, A to the power of six, which is not what we want either.

Here's some more questions for you to try, pause the video to complete your task, click resume once you're finished.

Let's take a look at part A first.

So the value of M must be three, as three subtract five would give us negative two.

For part B, we have a value of negative two.

This is because five subtract negative two would give us positive seven, which is what we wanted.

For part C, the power of M is actually five, because five subtract three is two, and our coefficient P is the value of three.

This is because 15 divided by three would give us a new coefficient of five.

And for our last example, we have a value of 12 for M, as 12 divided by four would give us our coefficient of three, and a power of P would be positive seven.

And again this is because four subtract seven, would give us negative three.

Pause the video to complete your task, click resume once you're finished.

And here is the solution.

So the question asks us to write an expression, in terms of A, for the average speed of the boat.

Well we're given two pieces of information, we're told that the boat travels, 35 A to the power of six kilometres.

And we're told that it does this in the time of seven A to the power of nine hours.

So therefore, to work out the speed, we're going to work out the distance divided by the time.

So we need to calculate 35 A to the power of six, divided by seven A to the power of nine, which would eventually give us our solution of five A to the power of negative three kilometres per hour as an average speed.

And that wraps up today's lesson, I hope you've had fun with this, and I hope you're feeling more confident with the division law involving indices.

I will hopefully see you soon.