Lesson video

Hi, everyone.

My name is Ms. Coo.

And I'm really happy that you're learning with me today.

We are going to be looking at Arithmetic Procedures, Index Laws, a really interesting lesson.

I hope you enjoy it.

I know I will.

So let's make a start.

Hi, everyone, and welcome to this lesson on "Problem solving with the laws of indices" under the unit Arithmetic Procedures, Index Laws.

And by the end of the lesson, you'll be able to use your knowledge of the laws of indices to solve problems. So let's have a look at some keywords.

16 is the fourth power of two.

Alternatively, this can be written as two with a little floating four to the right, which is read as two to the power of four.

A vinculum is a horizontal line placed over an expression to show that everything below that line is one group.

And numbers that have been multiplied by themselves as a repeated number of times can be expressed using a base and an exponent.

For example, two multiplied by two multiplied by two is written as two to the power of three.

We know the three is the integer exponent and identifies how many times the base has been multiplied by itself.

And the two.

Well, the two is the base.

Well, this represents the base and is the number or sometimes term that's been multiplied by itself.

An alternative word for exponent is index and you'll be hearing this word index or indices a lot through the lesson.

Now today's lesson will be broken into two parts.

We will look at simplifying using laws of indices and then solving using laws of indices.

So let's make a start.

Simplifying using laws of indices.

Now here are our laws of indices summarised.

Using our multiplication law, we know that when the bases are the same and multiplying, we sum the exponents.

We know when the bases are the same and we're dividing, we subtract our exponents.

And when we have a raised power, we multiply our exponents.

Also remember, a to the power of negative m basically means the reciprocal of the base and the exponent.

And a to the power of one over m means the mth root of a.

As such, problem-solving questions can involve a combination of more than one of these laws.

So combining these laws, let's have a look an example question.

Here, we're asked to express the following as a power of five.

We have five to the power of eight multiplied by five squared, all divided by five to the power of four.

We need to have a think.

Which laws do you think we're using here? Well, we're using the multiplication law.

In other words, where we have the bases the same, we add the exponents and we're using the division law.

In other words, where the bases are the same, we subtract the exponents.

So let's apply them.

Well, let's use the multiplication law first.

We have five to the power of eight multiplied by five squared.

This gives us five to the power of 10.

Then, we're going to use our division law next because we have five to the power of 10 divided by five to the power of four.

So we subtract those indices.

10 subtract four gives me a final answer of five to the power of six.

So therefore, our answer is five to the power of six.

We've expressed the calculation as a power of five.

So now what I want you to do is a quick check.

I want you to simplify and give the answer as the power stated in the question.

See if you can give it a go.

Press pause for more time.

Well done.

Let's see how you got on.

Well, for A, you should have summed the exponents of five and nine to give you 14.

So therefore, we have four to the 14 divided by four to the seven.

Subtracting those exponents, we have four to the power of seven.

For B, summing the exponents of five and six, we have two to the power of 11, then divided by two to the power of eight.

Because the bases are the same, we can subtract.

So that means it's 11 subtract eight, which gives us three.

So our answer as a power of two is two to the power of three.

Next, we have a to the power of eight multiplied by a to the power of 10, all divided by a to the five multiplied by a squared.

So let's apply those laws of indices.

That means we have a to the 18 divided by a to the seven.

Our bases are the same so we simply subtract our exponents to give us a to the 11.

For D, same again.

We sum those exponents because we're multiplying and the bases are the same.

So we have y to the 13 divided by y to the 18.

Then we subtract our exponents, given we're dividing and we have the bases the same.

So it's y to the power of negative five.

Huge well done if you got this one right.

Problem-solving questions can involve more than two laws.

For example, this question wants us to express the following as a power of seven.

Which laws do you think we're using here? Well, hopefully you spotted we're using three laws in order to tackle this question.

First of all, we're going to use the law where we're raising the power.

So we're going to rewrite seven squared, all to the power of three, as a single power of seven.

This would give me seven to the power of six.

Then I still have my multiplication of seven to the eight, all divided by seven to the four times seven squared.

Then using multiplication law, we can sum those exponents to give me seven to the power of six, add eight, over seven to the power of four, add two.

This then gives me seven to the 14, all divided by seven to the six.

Finally, we can use the division law, which tells us to subtract our exponents.

14 subtract eight gives me the answer of seven to the eight.

So that means our answer is seven to the power of eight, and we've expressed our answer as a power of seven.

Now what I want you to do is have a look at this working out.

Can you spot the mistake in this working out? And I also want you to work out the correct answer.

See if you can give it a go.

Press pause if you need more time.

Great work.

Let's see how you got on.

Well, given that the fact that the bases the same, when applying the law of division, there needs to be a subtraction of minus four from seven.

So that means the final last step should have been of seven subtract the negative four, which gives us five to the 11.

So the correct answer would be five to the power of 11.

Well done if you got this.

Great work, everybody.

So now it's time for your task.

What I want you to do is write the following as a power of two.

See if you can give this a go.

Show your working out and take your time.

Great work.

Let's have a look at question two.

Explain why Lucas is correct.

He says, all of the following will give a positive number.

See if you can give it a go.

Press pause for more time.

Great work, everybody.

Let's move on to these answers.

Here is our wonderful working out.

I want you to have a look at it and mark it.

Press pause, as you certainly need more time just to check.

Well done.

Let's have a look at question two.

Can you explain why Lucas is correct? Well, for the first one, using our laws of indices, this gives us minus four all squared.

Remember, any number squared always gives a positive number.

Well, for B, remember that negative exponent means we reciprocate.

It does not make the number negative.

And for C, everything in the brackets is to the power of zero.

So therefore, it equals one.

Well done if you got this.

Great work, everybody.

So now it's time for the second part of your lesson where we're solving using laws of indices.

Now other times, there is a need to form equations using laws of indices in order to solve for the unknown.

Here's an example of how an equation can be formed.

What we're asked to do is work out x.

We have seven to the power of 2x multiplied by seven to the power of nine gives us seven to the power of 19.

Now given all the bases are the same, we can form the equation just looking at the exponents.

Now remember, using the laws of indices, the multiplication law states that we add the exponents.

So all I've done is get the exponent of 2x, add it to the exponent of nine, and equated it all to that exponent of 19 because all of our bases of seven are the same.

Now from here, we can solve.

Subtracting nine from both sides gives us 2x is equal to 10.

Solving for x gives us x equals five.

Let's have a look at another example.

Here, we have y to the power of four multiplied by y to the power of eight, all divided by y to the power of x.

And it equals y squared.

So let's use these laws of indices again.

Here, we know we have the multiplication law, and the multiplication law states that we add the exponents.

Because of this vinculum, because of this dividing line, then we know we're subtracting the exponent of x.

So I've simply written four add x, subtract the x will give us the exponent of two.

And because all the bases are the same, that's why we can equate them.

From here, we can solve.

Well, the four add the eight is the 12.

12 subtract x equals two.

Rearranging and solving, we now know x equals 10.

So now what I'd like you to do is I'd like you to do a check question.

We have all the bases are the same, all the bases are 12, and I want you to work out the value of x, using your knowledge on laws of indices.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, we know using the multiplication law, we're summing those powers, x add nine.

Then we have a vinculum telling us to divide.

So we subtract the four, forming an equation.

x add nine subtract four equates to that exponent of 20.

From here, we can solve.

So therefore, x equals 15.

Huge well done if you got this one right.

So when there is a vinculum and we have bases the same throughout, it is good to write the equation separating the exponents which are above the vinculum and those exponents below.

For example, here we have three to the power of five, multiplied by three to the power of x, all over three squared multiplied by three to the four, and it gives us an answer of three to the power of 12.

Now once again, we have all the bases are the same, so that means we can just concentrate on those exponents only.

Using the laws above the vinculum, you can see the multiplication law states we add, so that's why it's five add our x.

We're subtracting because obviously we're dividing.

And then we have the laws below the vinculum.

Notice how we have three squared times three to the four.

Using our multiplication law, this means we add our exponents.

So it's indicated in those brackets, two add four.

This then equates to the exponent of 12.

Now we can solve.

So working this out, five add x.

I've just left it there.

The subtraction of the brackets.

Working out, two add four is six.

So I have five add x subtract six equals 12.

Simplifying a little bit more is the same as x subtract one equals 12, so therefore, x equals 13.

Now what I'd like you to do is a quick check question.

Take your time.

I want you to fill in the blanks to form and solve an equation.

See if you can give it a go.

Press pause for more time.

Well done.

Let's see how you got on.

Well, remember the laws of indices.

As we have the bases the same, we're multiplying those exponents.

Three multiplied by y is 3y.

So that must equal eight to the power of 12.

Equating the exponents because our bases are the same, that means 3y must be equal to 12.

So we can solve for y, working out y to be four.

Well done if you got this.

For B, we've got four to the power of x multiplied by four to the power of nine is equal to four to the power of 20.

We have bases all the same, so we can just focus on those exponents.

x add our nine must equal 20 because of the laws of multiplication.

Solving, we know x is equal to 11.

Let's have a look at a harder check now.

I want you to fill in the blanks to form and solve an equation.

See if you can give it a go.

Press pause if you need more time.

Well done.

Let's see how you got on.

Well, let's focus above and below that vinculum.

For A, you might notice all our bases the same.

So above the vinculum, we just have five to the 2x multiplied by five to the seven.

Focusing on those exponents and using our laws of multiplication, this is 2x add our seven subtract the three.

Because we're dividing by five to the three.

All equals that exponent of 18.

Now I can simplify a bit.

2x add seven and then subtract three simply gives me 2x add four.

The working out then moves on to 2x equals 14.

So x must equal seven.

For B, we have all the bases are nine.

So remember, look above and below that vinculum and concentrate on what's in the brackets.

Well, above the vinculum should be the summation of three and two.

Below the vinculum, we should simply have the x take away four.

Now let's simplify.

Three add two is our five.

Subtract.

Now expanding that bracket out, we should have the x add the four.

Remember, the negative multiplied by the x gives us negative x and negative multiplied by that negative four gives me plus four.

That was quite tricky.

Summarising gives me nine subtract x.

So therefore, x must equal negative one.

That was a really hard one.

Massive well done if you got that.

Now it's time for your task.

I want you to solve the following.

Take your time.

Press pause if you need.

Great work.

Let's move on to question two.

I want you to fill in the blanks to complete the working out.

Take your time.

Press pause if you need.

Great work.

Let's go through question three.

Question three wants you to form and solve an equation to find the value of x.

Take your time and press pause if you need.

Fantastic work, everybody.

Let's have a look at these answers.

I'm going to form my equation first.

So we should have x plus eight subtract three is 13.

So x plus five equals 13, giving me the value of x to be eight.

For B, 4y equals 20, so y equals five.

And for C, 2x add three in brackets subtract eight equals 19.

So therefore, 2x subtract five gives 19.

Solving for x gives me 12.

x is equal to 12.

Really well done if you got question one right.

For question two, here's our missing information.

Huge well done if you got this one right.

This was quite a helpful question, as it structured it for you.

For question three, this is a lot harder, as you haven't got that reliance of the structuring of the work like before.

So we know all the bases are the same.

Let's focus above and below that vinculum.

We should have eight add our 2x add four, then subtract the six.

Solving, we should have x is equal to 12.

For B, we have to apply our laws of indices first.

So simplifying, you should have two to the power of 24, multiplied by two to the power of 3x add six, all over two to the power of four.

Then, focusing on those exponents, we have this equation which solves to give x is equal to negative four.

That was a really tough question.

Massive well done if you got this one right.

Great work, everybody.

So problem-solving questions with index laws combines more than one law, and sometimes, the question wants the calculations to be simplified, and other times, it wants you to solve an equation using indices.

And when simplifying, ensure you know which laws to apply.

We know the multiplication law means when the bases are the same, you sum the exponents.

The division law means when the bases are the same, you subtract the exponents.

Raising the power means you multiply the exponents.

When you have a to the power of negative m, it means the reciprocal of a to the m.

And a to the power of one over m means the mth root of a.

Then solving, remember you can form equations using these laws so to solve more efficiently.

For example, the question wants us to solve eight to power of three, all to the power of y, which equals eight to the 12.

This is the same as eight to the power of 3y, which is equal to eight to the 12.

Given our bases are the same, we can focus on the exponents.

So 3y equals 12, giving us y is equal to four.

Great work today, everybody.

It was a tough lesson.

Massive well done.