Lesson video

Hi everyone, my name is Ms Coo, 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'll be able to use your knowledge of the laws of indices to solve problems. Let's 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 one.

For example, five and one fifth, these are reciprocals of one another because five multiply by one fifth does equal one.

Two-thirds and three over two, these are reciprocals of one another because two-thirds multiply by three over two does equal one.

A non example is four and 0.

4.

This is because four multiply by 0.

4 does not equal one.

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 one over or the reciprocal of the base and the exponent.

A to the power of one over M, means the nth 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 five.

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 as 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 one over m is the mth route of a, and you can see that in the question.

So I'm going to replace that Q route with the power of one third.

Then I've spotted the law of multiplication and division, so simplifying five to the power three multiplied by five squared gives me 5/20, all over five to the power of five, then this is all to the power of a third.

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

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

15 multiplied by the exponent of third gives me five to the power of five.

So that means using a range of laws, we've written our answer as a power of five, our answer is five to the power of five.

Now what I'd like you to do is a check question, I want you to express the following as a power of two.

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 a half.

Next I'm going to use my laws of indices and simplify 2 to the power of 78, multiply 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 a half.

This then gives me 2 to the power of 68, all to the power of a half, 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: eight to the power of negative two thirds.

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

So let's, let's apply them.

First of all, recognise that negative index means we need to reciprocate, so this means 1/8 to the power of two thirds.

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

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

Well, I can work this out to be the cube root of eight is two, and then I'm going to square that to give me the answer of one over four.

So therefore eight to the power of negative two thirds is equal to a quarter.

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 you can give it a go, press pause for more time.

Let's see how you got on, well, we're gonna concentrate on Lucas's 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 two thirds into a unit fraction or squared gives me 64 to the power of a third or squared.

While I know the 64 to power a third means the cube root of 64, which I then need to square.

This gives me one over four squared giving me a final answer of one 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.

Eight to the power of negative m over n.

Have a little think.

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

You could also write it as one over the nth route of a 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 25/16 to give me 16/25.

And then we have a denominator of two, which means we square root, square root in the 16, square root in the 25, all to the power three, this gives us four over five 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 64/9, all to the power of negative one half.

And then we have 125/8 all to the power of negative two thirds.

So 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 64/9 gives us 9/64.

This then is all to the power of a half.

The power of a half means the square root, so we square to nine and then square it to 64 to give us three over eight.

Well done, if you have you got this.

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

Then let's work out the cube root of eight and the cube root of 125 all to the power of two, which equates to two squared over five squared, which is 4/25.

Well done, have 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.

Write the following as a power three, press pause for more time.

Well done, let's have a look at question three.

Take your time and evaluate the following: Remember your priority of operations as well.

Press pause for more time.

Well done! Let's have a look at question four.

Write the correct exponent to make the number in the centre, take your time when you're doing these questions, this one's quite tricky.

Well done! Let's move on to these answers, for a, you should have got 1/125 and for b, 1/81, 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.

This was a tricky one here, but you should have got a final answer of three to the 12th.

Great work, press pause if you need.

For question three, 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.

For question four, this was incredibly hard.

Fantastic work! Have you got any of these? Great work everybody, so now let's have a look at problems involving different bases.

So here's a question, we have two to the power of 10 multiply by eight to the power of two, and the question wants us to write the answer as a single power of two 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 bases both two." So how do you think we can convert the base of eight into a base of two? Well we can use the laws of indices again, eight to the power of m, all to the power of n is equal to eight to the power of m multiplied by n, and this helps us change bases.

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

So here's my eight and then there's my squared, so my calculation becomes two to the power of 10 multiplied by two cubed all squared.

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

Now we have all the bases of two, so we can apply the multiplication law to give us two to the power of 16.

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

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

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

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

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

So we've calculated the answer to a power of three, nine to the power of 10, multiplied by three to the power of five gives me three to the power of 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 five.

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 five cubed.

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

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

I can use my laws of indices to work out my calculation to be five to the six times five to the eight times five cubed, and then I can work out my final answer to be five to the 17.

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

A hundred and twenty five squared times 25 to the four times five cubed is five to the para 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 four times four squared, all divided by two to the five is equal to two to the x.

From this question it is clear that there is a need to change all the basis to two 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 two, have a little think.

While changing all the bases to the two, we have two to the four is exactly the same as our 16, two squared is exactly the same as the four.

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

So now I have a new equation, two to the four, all to the four, multiply by two squared, all squared divided by two to the five which equals two to the x.

Now I'm going to apply my laws of indices, so I'm going to have two to the 16, multiply by two to the four, all divided by two to the five is two to the x.

Now remember laws of indices again using the multiplication and division law.

I can simplify this to give me two to the 20 divided by two to the five, which is two to the x.

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

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

Here we've got an example question, another example where we're asked to work out x.

We have y to the four multiply by y to the eight 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 four add eight 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.

Four, add eight, subtract the x gives you the exponent two, 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 of four 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 four, so that means using our laws and multiplication, we sum our 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 eight.

This was a really good question and it's important to 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, two to the 14 is equal to four to the power of 12, multiplied by eight to the power of x plus two all divided by two to the four.

First they were 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 two, because we can change four to a base of two, eight to a base of two.

So that means four is equal to two squared, eight is equal to two cubed.

Now we can apply our laws of indices to simplify the equation further, giving me two to the power 24, multiplied by two to the power 3x plus six all divided by two to the four.

Using the laws of indices, remember that multiplication laws states that we can add the exponents.

Now our base is the same, this means focusing on our exponents above the vinculum, the exponents are 24 and 3x plus six.

Because we're multiplying, that means we add those exponents using the laws of division, that means we subtract the exponents, so that's why it's a subtraction of four.

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 b negative four.

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.

See if 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.

So you can give it a go, press pause one 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 you 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 five is your best choice, then use your laws of indices to solve for x where x is equal to negative 11/9.

Press pause for more time to mark, well done.

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

So using our laws of indices form an equation with those exponents and you would've found x to b negative one third.

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.

And 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 nine and for b, the coefficient of x is negative seven.

And for c, the coefficient of x is negative 39/10.

This is a nicer 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, multiply by 10x to the four equals 18x to the six.

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 exponent of six is five, but we do not add the coefficients.

Eight add 10 yes equals 18, but in the context of the question, we are multiplying 8x squared by 10x to the four that is not 18x to the six.

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

Well the correct answer is 80x to the six, because the multiplication law states that we add the exponents two and four does equal the six and we multiply the coefficients given the question states: Eight multiply 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 one 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 four times two is eight, which is correct.

However, they wrongly applied this law to the coefficients.

What they should have done is worked up three squared, which is nine, so the correct answer is nine x to the eight.

B is incorrect, but why is b incorrect? Well, using the law's division, the correct and subtracting three from 10, which gives us seven.

However, they've wrongly applied the law to the coefficients.

We should have done 40 divided by eight, which gives us five.

So the correct answer is 5x to the power of seven 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 four as a result of x squared all squared, but x the power of four, all cubed gives x of the 12, not X of the 64.

Then using the laws of division, the exponent must be subtracted and they have wrongly multiplied the coefficients also.

So let's work it out correctly.

Well 4x squared all squared gives the 16x of the four, 3x of the four all cubed is 27x of the 12.

Remember three cubed is 27, all diverted by 2x of the four.

Then from here we simply divide 432 by 2 to give us 216.

X the power of 16 divided by x of the four gives us x to the power of 12.

So recognising the laws of indices only applies to exponents, allows us to correctly form and solve equations to, for example, this question wants us to solve to find the positive value for a.

We have three a all squared is equal to nine 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 gonna isolate the a term by dividing everything by nine.

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

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

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

From here we can identify a to b nine.

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 two, multiplied by 2x cubed all to the power of three, and that equals 2px, the six all squared, so let's our coefficients and our exponents.

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

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

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

We know the coefficients of x, the 12th 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 five making all the terms have the same base allows me to correctly use those laws of indices.

So that means I have five of the power of five all cubed, five to the power of three, all to the power of six, and that still equals five to the power of 2x plus one.

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

So now I can equate my exponents, this means 15, subtract 18 is equal to the exponent of 2x at one.

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

Really well done if you got this.

For b, much the same, I'm going to convert each term so I have the same base of three, then from here I'm going to equate those exponents.

Above the vinculum, I have eight and my 2x plus four, and then I have a subtraction of six, 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, have you got this? Well, the nth root of axe the 9th is equal to 2x cubed.

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

So we have axe of the 9th alter the power of 1/n, which is 2x cubed.

Using our laws, I'm just going to multiply that nine by the 1/n to give me x of the power of 9/n, and then I know I'm working out that nth route 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/n must equal x cubed.

So that means we know n has to be three.

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

The cuberoot of a must give me the coefficient of two, so that means a must be eight.

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 two to the power of 10, multiply by eight squared to a single power of two.

This means we change all the bases to two.

In other words, two to the 10, multiply by two cubed, all squared is equal to two to the 10, multiply by two to the six, which is two to the 16th.

Another example would be where we have to solve for x, we have five to the x all over five to the 4x minus two, which equals five cubed.

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

This was a really tough lesson, a massive well done.

It was great learning with you.