Lesson video

Hi, everyone.

My name is Miss Ku.

And today we're looking at Arithmetic Procedures: Index Laws.

I really hope you enjoy the lesson.

It's going to be challenging.

It's got lots of fun tasks to do.

I know I'm going to enjoy teaching it.

So let's make a start.

Hi, everyone and welcome to this lesson on advanced problem solving with the laws of indices and it's under the unit, Arithmetic Procedures: Index Laws.

And by the end of the lesson you will be able to use your knowledge of the laws of indices to solve problems. Let's a look at some keywords.

Now remember, reciprocal is the multiplicative inverse of any non-zero number.

And any non-zero number multiplied by its reciprocal is equal to 1.

For example, 5 and 1/5, these are reciprocals of one another because 5 multiplied by 1/5 does equal 1.

2/3 and 3/2, these are reciprocals of one another because 2/3 multiplied by 3/2 does equal 1.

A non example is 4 and 0.

4.

This is because 4 multiplied 0.

4 does not equal 1.

Today's lesson will be broken into three parts.

We'll be looking at fractional and negative indices first, then problems involving different bases and then harder problems using index laws.

So let's make a start, looking at fractional and negative indices.

Recapping on our laws of indices, remember the law of multiplication states, when the basis are the same, we add our exponents.

Laws of division, it states that when the basis are the same, we subtract our exponents.

Raising to the power means we multiply our exponents.

When you have a base to a negative exponent, that means it's 1 over or the reciprocal of the base and the exponent.

A to the power of 1 over m means the mth root of a, and a to the power of m over n means the nth root of a all to the power of m.

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

So let's have a look at this question.

We're asked to express the following as a power of 5.

What I want you to do is think about which laws do you think we're using here.

Have another think.

Well, we're using four laws here.

It's quite a lot in one question.

So let's break this question down according to each law.

Well first things first, we've got a to the power of 1 over m is the mth root of a, and you can see that in the question.

So I'm going to replace that cube root with the power of 1/3.

Then I've spotted the law of multiplication and division.

So simplifying 5 to the power 3 multiplied by 5 squared gives me 5 over 20, all over 5 to the power of 5.

Then this is all to the power of 1/3.

Now I'm going to use my law of division and I'm going to rewrite 5 to the 20 divided by 5 to the 5 as 5 to the 15.

Now I'm going to use my final law to simplify further.

15 multiplied by the exponent of 1/3 gives me 5 to the power of 5.

So that means using a range of laws, we've written our answer as a power of 5.

Our answer is 5 to the power of 5.

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

I want you to express the following as a power of 2.

Take your time.

Press pause if you need.

Well done.

Let's see how you got on.

Well, first things first, hopefully you recognise that square root is exactly the same as the power of 1/2.

Next I'm going to use my laws of indices and simplify 2 to the power of 78, multiplied by 2 to the power of 40 as 2 to the power of 118, and this is all over 2 to the power of 50, which is then raised to the power of 1/2.

This then gives me 2 to the power of 68, all to the power of 1/2, which gives me 2 to the power of 34.

Really well done if you've got this.

Now what we're going to do is combine two laws.

Here, we've got an example question.

8 to the power of negative 2/3.

So which laws do you think we're using here? Well, we're using a to the power of negative m, which is the reciprocal, the base and the exponent and a to the power of m over n, which means the nth root of a all to the m.

So let's apply them.

First of all, recognise that negative index means we need to reciprocate.

So this means 1 over 8 to the power of 2/3.

Then we know using the next law, we can rewrite this using roots and exponents.

So this means one over the cube root of 8 all squared.

Well, I can work this out to be the cube root of 8 is 2 and then I'm going to square that to give me the answer of 1 over 4.

So therefore, 8 to the power of negative 2/3 is equal to 1/4.

What I want you to do is have a look at a quick check.

Here are some calculations from Lucas and Andeep.

I want you to have a look at them and identify their mistakes and then I'd like you to work out the correct answer.

See if we can give it a go.

Press pause one more time.

Let's see how you got on.

Well, we're gonna concentrate on Lucas' first.

First of all, Lucas has incorrectly interpreted that negative exponent.

The negative exponent does not make the number negative.

It reciprocates the number.

And for Andeep, well, Andeep has correctly identified that the negative exponent means to reciprocate the number, but the denominator of the fraction indicates the cube root of 64 and then the numerator indicates to square the result.

So let's work out the correct answer.

The negative exponent simply identifies that we reciprocate the number and the exponents.

From here, rewriting the power of 2/3 into a unit fraction all squared gives me 64 to the power of 1/3 all squared.

Well, I know the 64 to power of 1/3 means the cube root of 64, which I then need to square.

This gives me 1 over 4 squared giving me a final answer of 1 over 16.

Massive well done if you got this one right.

So I want you to have a look at this, and I want you to think what does this summarise to, 8 to the power of negative m over n.

Have a little think.

Well, it means one over the nth root of a all to the power of m.

You could also write it as 1 over the nth root of a all to the m.

Either one of those are absolutely fine.

So now what I want you to do is, using this law, which of the following is correct? Have a little think.

Press pause one more time.

Well, you should have discovered a is not correct but b is correct.

Remember, the negative exponent does not mean we put a negative in front of the number.

The negative exponent means we reciprocate.

This means B is correct.

The negative exponent means we reciprocate the 25 over 16 to give me 16 over 25.

And then we have a denominator of 2, which means we square root, square root in the 16, square root in the 25 all to the power 3.

This gives us 4 over 5 all to the power 3, 64 over 125.

Well done if you've got this one right.

Now, it's time for another check.

I want you to take your time and evaluate the following.

We have 64 over 9 all to the power of negative 1/2.

And then we have 125 over 8 all to the power of negative 2/3.

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, remember that negative exponent means we reciprocate.

So reciprocating 64 over 9, gives us 9 over 64.

This then is all to the power of 1/2.

The power of 1/2 means the square root.

So we square root a 9 and then square it of 64 to give us 3 over 8.

Well done if you got this.

For b, that negative exponent means we reciprocate, so that means we have 8 over 125 all to the power of 2/3.

Then let's work out the cube root of 8 and the cube of 125 all to the power of 2, which equates to 2 squared over 5 squared, which is 4 over 25.

Well done if you got this one right.

Great work everybody.

Now it's time for your task.

I want you to evaluate the following.

Press pause for more time.

Well done.

Let's have a look at question two.

Take your time and evaluate the following.

Remember your priority of operations as well.

Press pause for more time.

Let's move on to these answers.

For a, you should have got 1 over 125 and for B 1 over 81.

Press pause if you need to have a look at this working out and mark.

The continuation of question one.

Here's the working out, and here are our answers.

Well done if you've got these.

Press pause to compare and mark.

Well done.

Let's have a look at question two.

We should have got this type of working out and then here are our answers.

Really well done.

This was a tricky one.

Press pause to have a look and mark.

Great work, everybody.

So now let's have a look at problems involving different bases.

So here's a question.

We have 2 to the power 10 multiplied by 8 to the power 2.

And the question wants us to write the answer as a single power of 2 with an integer exponent.

Izzy says, "This is impossible as they have different bases." But Sofia says, "This is possible because we can make the base both 2." So how do you think we can convert the base of 8 into a base of 2? Well, we can use the laws of indices again.

A to the power of m all to the power of n is equal to a to the power of m multiplied by n.

And this helps us change bases.

In other words, we can change 8 into a base of 2 because we know 2 cubed is equal to 8.

So here's my 8 and then there's my squared.

So my calculation becomes 2 to the power of 10 multiplied by 2 cubed all squared.

Now we can apply this law and work out the calculation to be 2 to the power of 10 multiplied by 2 to the power of 6.

Now we have all the bases are 2.

So we can apply the multiplication law to give us 2 to the power of 16.

So that means we've calculated the answer as a power of 2 with an integer exponent.

So now I'm going to do a question and then I'd like you to do the next question here.

It wants us to write the answer as a single power of 3.

Well, first things first, I recognise that 9 can be written as a base of 3 because 3 squared is exactly the same as 9.

So you can see I've replaced that 9 with 3 squared and it's still to the power of 10.

Now I can use my laws of indices and I can work out that 3 squared all to the power of 10 is equal to 3 to the power of 20, which is then being multiplied by 3 to the 5.

And then using my laws of indices, I've worked out my final answer to be 3 to the power of 25.

So we've calculated the answer to a power of 3.

9 to the power of 10 multiplied by 3 to the power 5 gives me 3 to the power 25.

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

Here, the question wants you to write the answer as a single power of 5.

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, I'm going to replace that 125 with 5 cubed.

I'm gonna replace that 25 with 5 squared and 5 cubed just stays.

Now I have every single term as a power of 5.

I can use my laws of indices to work out my calculation to be 5 to the 6 times 5 to the 8 times 5 cubed, and then I can work up my final answer to be 5 to the 17.

So that means we've calculated the answer as a power of 5.

125 squared times 25 to the 4 times 5 cubed is 5 to the power of 17.

Really well done if you've got this.

But sometimes the question doesn't explicitly tell you to change the base.

For example, here it wants you to find the value of X and show all algebraic workings.

We have 16 to the 4 times 4 squared, all divided by 2 to the 5 is equal to 2 to the x.

From this question, it is clear that there is a need to change all the bases to 2 and then equate those exponents.

So what I want you to do is, I want you to have a look at each term and can you change each power to a base of 2? Have a little think.

Well changing all the bases to the 2, we have 2 to the 4 is exactly the same as our 16, 2 squared is exactly the same as the 4.

So all I've done is replaced that 16 with 2 to the 4 and the 4 with 2 squared.

So now I have a new equation, 2 to the 4 all to the 4 multiplied 2 squared all squared divided by 2 to the 5 which equals 2 to the x.

Now I'm going to apply my laws of indices, so I'm going to have 2 to the 16, multiplied by 2 to the 4, all divided by 2 to the 5 is 2 to the x.

Now remember laws of indices again, using the multiplication and division law, I can simplify this to give me 2 to the 20 divided by 2 to the 5, which is 2 to the x.

And simplifying gives me 2 to the 15 is 2 to the x.

Given the basis are the same, that means we know x is equal to 15.

Here we've got an example question.

So an example where we're asked to work out x.

We have y to the 4 multiplied by y to the 8 over y to the x equals y squared.

Now using our laws of indices, the multiplication laws state that we add the exponents.

So that means it's 4 add 8.

And using the laws of division, we know we subtract the exponent, so we're subtracting the exponent of x.

And because we have all the bases are the same, now we can equate all of those exponents.

4 add 8 subtract the x gives you the exponent 2, which then can be solved where x is equal to 10.

So now what I want you to do is work out the value of x given the following equations.

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 for a, we have all the bases of the same.

So that means using the laws of multiplication, we sum those exponents, and then laws of division state we subtract the exponent to 4 and that equates the exponent to 20.

Solving gives me x is equal to 15.

For b, all the bases are the same.

They're all 4.

So that means using our laws and multiplication, we sum are exponents above the vinculum, and subtract the exponents below the vinculum, which then equate to 10.

Working this out, we can solve for x being 8.

This was a really good question and it's important to separate the exponents either side of the vinculum and form your equation from there.

Now let's apply it where we have different bases.

For example, 2 to the 14 is equal to 4 to the power of 12 multiplied by 8 to the power of x plus 2 all divided by 2 to the 4.

Firstly, we're going to change them to the same base.

Which base do you think we should use? Well, it's best to use the base of 2 because we can change 4 to a base of 2, 8 to a base of 2.

So that means 4 is equal to 2 squared, 8 is equal to 2 cubed.

Now we can apply our laws of indices to simplify the equation further, giving me 2 to the power 24 multiplied by 2 to the power 3x add 6 all divided by 2 to the 4.

Using the laws of indices, remember that multiplication law states that we can add the exponents now our bases are the same.

This means focusing on our exponents.

Above the vinculum, the exponents are 24 and 3x plus 6.

Because we're multiplying, that means we add those exponents.

Using the laws of division, that means we subtract the exponent.

So that's why it's a subtraction of 4.

Then it has to equate to the exponent of 14.

Because all the bases are the same, so we can equate those exponents.

Then solving, we can work out x to be negative 4.

This was a tough question as we're still forming equations to solve, but we had that extra layer of complexity by ensuring that we convert to the same base.

Great work, everybody.

So now what I want you to do, have a look at question one and write your answer according to the power stated.

So you can give it a go.

Press pause if you need more time.

Great work.

Let's have a look at question two.

Work out the exact value for x.

Look at those different bases and have an idea of what you need to do.

See if you can give it a go.

Press pause for more time.

Well done.

Let's move on to question three.

Same again, you need to work out the exact value for x.

Look at those bases.

You might need to change something before you start using those laws of indices.

See if can give it a go.

Press pause for more time.

Well done.

Let's go through these answers.

Well for question one, here's our wonderful working out.

And our answers, how did you get on? Press pause if you need more time to mark.

For question two, convert to the same base, the base of 5 is your best choice, then use your laws of indices to solve for x where X is equal to negative 11 over 9.

Press pause for more time to mark.

Well done.

For question three, we had to change them to a base of 5.

So using our laws of indices, form an equation with those exponents and you would've found x to be negative 1/3.

Really well done.

Great work, everybody.

Now, it's time to look at harder problems using index laws.

Now we've looked exclusively at powers.

So now it's important to consider the coefficient of a term.

Remember a coefficient in general is any of the factors of a term relative to a given factor of the term.

So let's recap and identify the coefficient of the x term.

Here, I've written some expressions and I want you to identify the coefficient of the x term.

Press pause if you need a little bit more time.

Well for a, the coefficient of x is 9 and for b, the coefficient of x is negative 7, and for c, the coefficient of x is negative 39 over 10.

This is a nice little recap on what coefficient is.

It's important to remember, the laws of indices only apply to the exponents or the indices.

We do not apply these laws to the coefficients.

For example, let's have a look at this question.

Somebody has worked this out and said, 8x squared multiplied 10x to the 4 equals 18x to the 6.

Can you spot and explain a common mistake? Well, hopefully you're spotted that the multiplication law does state that we add the exponents because we have the same base which is x.

So the exponents of 6 is fine, but we do not add the coefficients.

8, add 10 yes equals 18.

But in the context of the question, we are multiplying 8x squared by 10x to the 4.

That is not 18x to the 6.

So let's see if we can work out the correct answer.

Well, the correct answer is 80 x the 6 because the multiplication law states that we add the exponents, 2 and 4 does equal the 6 and we multiply the coefficients, given the question states 8 multi[Lied by 10, which is 80.

So what I want you to do for this check question is to identify the mistakes in each of the following and work out the correct answer.

See if you give it a go.

Press pause for more time.

Well done.

Let's see how you got on.

Well, this is incorrect.

Raising the index to a power, we multiply the exponent.

So 4 times 2 is 8, which is correct.

However, they wrongly applied this law to the coefficients.

What they should have done is worked up 3 squared, which is 9.

So the correct answer is 9x to the 8.

B is incorrect.

But why is b incorrect? Well, using the law's division, they're correct in subtracting 3 from 10, which gives us 7, however they've wrongly applied the law to the coefficients.

We should have done 40 divided by 8, which gives us 5.

So the correct answer is 5x to the power of 7.

And for c, c is incorrect.

They've incorrectly used a range of indices here.

First of all, they correctly gave us x to the power of 4 as a result of x squared all squared, but x the power of 4 all cubed gives x to the 12, not x to the 64.

Then using the laws of division, the exponent must be subtracted.

And they wrongly multiplied the coefficients also.

So let's work it out correctly.

Well, 4x squared all squared gives a 16x of the 4, 3x of the 4 all cubed is 27x to the 12, remember, 3 cubed is 27, all divided by 2 x of the 4.

Then from here we simply divide 432 by 2 to give us 216, x to the power of 16 divided by x of the 4 gives us x the power of 12.

So recognising the laws of indices only applies to exponents, allows us to correctly form and solve equations too.

For example, this question wants us to solve, to find the positive value for a.

We have 3a all squared is equal to 9 cubed.

Well first of all, let's expand the coefficients and the exponents.

So this gives us 9a squared is equal to 9 cubed.

Now we're going to isolate the a term by dividing everything by 9.

So that means we have a squared is equal to 9 cubed over 9.

So that means we're going to apply another law of indices, which is the law of division giving us, 9 squared, because it's a result of 9 cubed divided by 9.

So that means we know a squared is equal to 9 squared.

From here we can identify a to be 9.

Now let's extend this further.

We can equate coefficients.

For example, we're gonna solve for p, where p is a positive number.

Well, I'm asked to give p correct to one decimal place.

Here we have 4x to the power of 1.

5 all to the power of 2, multiplied by 2x cubed all to the power of 3 and that equals 2px to the 6 all squared.

So let's expand our coefficients and our exponents.

Well this gives us 16 x cubed multiplied by 8x to the 9, which then equates to 4p squared x the 12.

Now using the laws of multiplication, we know 16 multiplied by 8 is 128, x cubed multiplied by x to the 9 is x to the 12 and this still equals 4p squared x to the 12.

Now let's equate the coefficients of x to the 12.

We know the coefficients of x to the 12 are 128 and 4p squared.

Now from here, we know p squared must equal 32.

So p is equal to the square to 32, which is 5.

7 to one decimal place.

Fantastic work, everybody.

So now let's have a look at your practise questions.

For question one, I want you to solve for x.

Take your time, look at those different bases and have a strategy in mind.

See if you can give it a go.

Press pause for more time.

Well done.

Let's have a look at question two.

Question two wants you to work out the value of a and n.

Take your time.

Press pause if you need, Great work.

So let's have a look at these really tough problem solving questions.

Well, for question two, all I'm going to do is change so I have the basis of 5.

Making all the terms have the same base allows me to correctly use those laws of indices.

So that means I have 5 to the power of 5 all cubed, 5 to the power of 3 all to the power of 6.

And that still equals 5 to the power of 2x plus 1.

Now simplifying using the laws of indices, we have 5 to the 15 over 5 to the 18.

So now I can equate my exponents.

This means 15 subtract 18 is equal to the exponent of 2x add one.

Then working this out I can solve for x, giving me x is equal to negative 2.

Really well done if you've got this.

For B, much the same.

I'm going to convert each term so I have the same base of 3.

Then from here I'm going to equate those exponents.

Above the vinculum, I have 8 and my 2x plus 4 and then I have a subtraction of 6, which gives me the exponent to 30.

Solving from here, I then have x equal to 12.

For question two, this was an incredibly hard question, so well done if you got this.

Well, the nth root of axe to the 9 is equal to 2x cubed.

First of all, remember the root means the exponent is a unit fraction where n is the denominator.

So we have axe to the 9 all to the power of 1 over n, which is 2x cubed.

Using our laws, I'm just going to multiply that 9 by the 1 over n to give me x of the power of 9 over n.

And then I know I'm working out that nth root of a, which still equals to x cubed.

So now let's equate the x terms. So equating the x terms we know x to the power of 9 over n must equal x cubed.

So that means we know n has to be 3.

Knowing n equals 3, that means we can substitute it back in.

The cube root of a must give me the coefficient of 2.

So that means a must be 8.

This was a super tough question.

Really well done if you figured this one out.

Fantastic 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.

Given the laws of indices only apply when the base is the same, it's sometimes required to change the bases.

For example, simplify 2 to the power of 10 multiplied by 8 squared to a single power of 2.

This means we change all the bases to 2.

In other words, 2 to the 10 multiplied by 2 cubed all squared is equal to 2 to the 10 multiplied by 2 to the 6, which is 2 to the 16.

Another example would be where we have to solve for x.

We have 5 to the x all over 5 to the 4x minus 2, which equals 5 cubed.

So equating those exponents, we can form an equation and then solve, x subtract 4x takeaway 2 equals 3 and then we can solve from there.

This was a really tough lesson.

A massive well done.

It was great learning with you.