### Video Transcript

In this video we’re gonna use all of our knowledge of powers, indices, and
exponents to simplify algebraic expressions into a series of terms in power form. This is
especially useful when you start learning about calculus and need the power form to
differentiate and integrate functions.

Before you attempt the questions in this video we expect,
you to know these things. We expect you to know the definition of 𝑥 to the power of
𝑎; we write down 𝑥 𝑎 times and multiply them all together. We expect you
to understand negative powers, which refer to reciprocals, and we also expect you to be able to
understand fractional exponents. So the denominator of the fraction tells you what root, and the
numerator of the fraction tells you what power. We also expect you to be able to apply the
addition, the subtraction, and the multiplication rule.

So our first question is to take this
expression and to rearrange it into a series of terms in power form. Now first of all, I’d split
this into two separate fractions. So in the first case, we’ve got the square root of 𝑥
over 𝑥 and then we’re adding one over the square root of 𝑥. And
straight away we can cancel out this; how many times does the square root of 𝑥 go
into the square root of 𝑥? It goes in once. How many times does it go into the square
root of 𝑥? It goes in once.

So that first term there, square root of 𝑥
divided by itself, is just one. Now on the second term, one over the square root of
𝑥, well the square root of 𝑥 that’s just a fractional power of a
half, so this is 𝑥 to the power of a half. So we kind of have got an answer there;
we have got some terms in power form. But we can tidy up that second term by just making it
𝑥 to the power of negative a half, so the reciprocal of what it is at the moment.

So remember 𝑥 to the power of negative a half, the negative means we do the
reciprocal of the base and the half tells us it’s square root, so we do have the same thing there.
So our next question, we’ve got three 𝑥 squared take away two lots
of the cube root of 𝑥 all over the fourth root of 𝑥, and we’ve got to
express that as a set of terms in power form.

So like before, I think I’d split that big
fraction up into two separate fractions first of all. So we’ve split them out there, so three
𝑥 squared over the fourth root of 𝑥 take away two lots of the cube
root of 𝑥 over the fourth root of 𝑥.

Now we can start converting some
of those individual 𝑥 terms into power format, these cubes- cube roots and
fourth roots. So the cube root of 𝑥 is 𝑥 to the power of a third and
the fourth root of 𝑥 is 𝑥 to the power of a quarter, so we can
rewrite that expression in this way.

So now I’m gonna take each term individually and think
about the first term is three 𝑥 squared divided by 𝑥 to the power of
a quarter, and the second term is two 𝑥 to the power of a third divided by
𝑥 to the power of a quarter. And then we can use our subtraction rule because we
got 𝑥 squared divided by 𝑥 to a quarter so that’s- we’re gonna
subtract the powers, and we got 𝑥 to the power of a third divided by
𝑥 to the power of a quarter so we can subtract those powers as well.

So remember
we have the same base, 𝑥, so we can just simply subtract those powers, two minus a
quarter. I’ve got the same base here, which means that we can subtract these powers, a third minus
a quarter. And two take away a quarter is one and three-quarters, and a third take away a
quarter is a twelfth. So we’ve got three 𝑥 to the power of one and three quarters
minus two 𝑥 to the power of a twelfth.

So we do have a series of terms in power
form, but we don’t tend to leave our powers as mixed numbers like this so I’m going still to
need to turn that into a top heavy fraction. And when I do that, I’ve now got, so one and three
quarter this is the same as seven over four, so that’s the answer: three lots of 𝑥
to the power of seven over four minus two lots of 𝑥 to the power of a twelfth.

So we’ve got our answer. But just to kinda go a step further, if we wanted to convert that back into
root format, three times 𝑥 to the power of seven over four; that four is the
fourth root; seven means to the power of seven. So that would be an equivalent expression for
that one.

And for the other one 𝑥 to the power of one over twelve means it’s the
twelfth root of 𝑥, so that would be the equivalent expression for that other one
there. As I say obviously, that was the answer to the question, but we’re just kind of trying to
extend your knowledge. In another question, you might have been asked something slightly different so
we just want to make sure that we went over again how to convert that back into the other
format.

Now we’ve got to simplify this horrible expression and put it into a series of terms in
power form. So stage one is going to be to replace all of these root terms with fractional
powers. And the cube root of 𝑥 is 𝑥 to the power of a third, and the
fifth root of 𝑥 is 𝑥 to the power of a fifth. But remember, we’ve got
one over the fifth root of 𝑥, so that brings a negative sign in there as well. It’s
gonna be 𝑥 to the power of negative a fifth.

Now the last term in the brackets
here. We’ve got 𝑥 to the power of negative a fifth all to the power of two. So
this we’re gonna be using our multiplication rule to simplify that term down. So
multiplying those powers together, negative a fifth times two is 𝑥 to the negative
two-fifths.

Okay, so now we can think about multiplying out these brackets term by term. And when
we’re doing that we’ve got the same base; it’s 𝑥 in each case, and we’ve got
different powers. So because we’re multiplying them, we’re going to be using the addition rule to
add those powers together.

So 𝑥 to the two-thirds times 𝑥 squared, add
those two powers together that’s 𝑥 to the two-thirds plus two; 𝑥 to
the two-thirds times 𝑥 to the third, add those two powers together, two-thirds plus
one-third; and then lastly, 𝑥 to the two-thirds lots of negative 𝑥 to
the negative two-fifths. Well it’s a positive term and a negative term, so that’s gonna give us a
negative term. Then we’re gonna do 𝑥; we’re going to add those powers two-thirds
add negative two-fifths.

So the first term there, two plus two-thirds, so we’ve got 𝑥
to the two and two-thirds. The second term, two-thirds plus-one third is one, so that’s
𝑥 to the power of one or just 𝑥. And two-thirds add negative two-
fifths, well when we work that all out it’s like two-thirds take away two-fifths; common
denominator of fifteen, we end up with four-fifteenths.

And the only problem with this is we’ve
got a mixed number as our power here, so we’re just gonna turn that into a top heavy fraction.
And there’s our answer: 𝑥 to the power of eight over three plus 𝑥
take away 𝑥 to the power four over fifteen. So that was our answer as we say, but
it’s worth practicing to make sure that you can convert into one format and then back into the
other format as well.

So just check that you understand how we got these answers down here.
And onto our next example, this is a bit of a beast, isn’t it? So what we gotta try and do with
this one is again come up with a series of expressions in index form, power form, and let’s
start breaking that down bit by bit.

So the first thing I’m gonna do is start attacking some
of these fourth roots and cube roots and so on and expressing those in index form or power
form. And the fourth root of 𝑥 squared, well the two goes on the top of the
fraction and the four goes on the bottom of the fraction, so that first bit here becomes
𝑥 cubed over 𝑥 to the two over four. Obviously that two over four is
gonna simplify to a half, but we’ll deal with that in a moment. And this bit here, so we’ve got
the cube root of 𝑥 to the power of four. So the cube root means there’s gonna be a
three on the denominator, and to the power of four means we’re gonna have a four on the
numerator of that.

And the last one I’m just gonna split it up in a slightly different way, just-
so we got two 𝑥 squared on the top. But because we’ve got the fifth root of
thirty-two 𝑥, I’m just gonna split those out into two separate terms: the fifth root
of thirty-two times the fifth root of 𝑥. And now I’m gon- I can work out what the
fifth root of thirty-two is, and the fifth root of 𝑥 is gonna be 𝑥 to
the power of one over five.

So we’ve simplified two over four to make it a half, and the
fifth root of thirty-two is two because two times two times two times two times two is
thirty-two. So that becomes a two, and the fifth root of 𝑥 is 𝑥 to the
power of a fifth. Right, now we’re gonna go through bit by bit; we’re gonna look at that term,
that term, and that term.

And we’ve got 𝑥 cubed divided by 𝑥 to the
power of a half, so we’re gonna use our subtraction rules for each of those individually
and just simplify those terms down. So the first bit, we’ve got 𝑥 cubed divided by
𝑥 to the half. And before I do the next bit, I’m just gonna split that three out
because that’s makes things a little bit difficult otherwise. So 𝑥 to the four
over three over three 𝑥 is the same as a third times 𝑥 to the four
over three over 𝑥.

I’m sure if you can multiply the tops out there and you
multiplied the bottoms out, you’d see that you get the same thing. But believe me, if we treat
this as a separate number on its own, now we can use our subtraction rule on this bit here
where we’ve got the same base; that’s gonna make our life a little bit easier. And then moving
on to the last term, well we’ve got a two on the top and a two on the bottom that both cancel out;
twos go into two once; twos go into two once; so we’re just left with 𝑥 squared
divided by 𝑥 to the power of a fifth.

Well now we can apply a subtraction rule to
this bit. We can turn this one into a division, and we can apply subtraction rule to this bit. So
picking them off term by term, 𝑥 cubed divided by 𝑥 to a half is
𝑥 to the power of three minus a half. That second term there, just rearranging that,
that’s 𝑥 to the four-thirds divided by 𝑥, which I’m gonna write as
𝑥 to the power of one because when we go to the next stage and do the
subtraction rule, it’s gonna be four over three take away one.

And then lastly on the end over
here 𝑥 squared divided by 𝑥 to the fifth, so we can subtract those
powers. That’s 𝑥 to the power of two take away a fifth. So three take away a half is
two and a half, so that first term is 𝑥 to the power of two and a half. And then
tidying up the next term, a third of 𝑥 to the power four-thirds minus one as we
said before. And then that last term, two minus fifth is one and four-fifths, so that becomes
𝑥 to the power of one and four-fifths.

So now we’re gonna need to tidy up some of
those powers a bit more; we don’t leave powers in mixed number format, remember? And
𝑥 to the power of two and a half, well two and a half is the same as five over
two. And four-thirds take away one, well one is three-thirds so four-thirds take away three-thirds
leaves one-third. And that last term, one and four-fifths is the same as nine over five.

So now
we’re ready to multiply out term by term and see what we get next. So multiplying that first
term by what’s outside the brackets, we’ve got 𝑥 to the five over two times a
third of 𝑥 to the power of a third. Well when you multiply things together, it
doesn’t matter what order you do that in, so I’m gonna bring the third out and put that first.
So we’ve got a third times 𝑥 to the power five over two times 𝑥 to the
power of a third. Now I’m gonna multiply the 𝑥 to the five over two by the next
term. So that’s 𝑥 to the power of five over two times 𝑥 to the power of
nine over five, taking away there.

And using the addition rule here, because we’re doing
𝑥 to the five over two times 𝑥 to the power of a third, so we’re going to
add those powers together. And the same for the second term there, five over two plus nine over
five is the power of 𝑥, and there we get 𝑥 to the power of forty-three
over ten. So that is our answer.

Now just before we go let’s play around with expressing
those in slightly different ways. So remember that this term here could also be written this
way, so a third of 𝑥 to the seventeen over six means a third times 𝑥
to the seventeen over six all over one. So on the top, we’ve got one times 𝑥, which
gives us just 𝑥 to the seventeen over six; and on the bottom, I’ve got three
times one, which is giving us the three.

And remember, the six on the denominator means it’s the
sixth root, and the seventeen on the numerator means it’s the seventeenth power, and that whole
thing is over three. Or we could’ve put the whole thing to the power of seventeen; it doesn’t
matter which of those two you use. And looking at the other expression, the ten on the
denominator means it’s the tenth root, and the forty-three on the numerator means it’s the
forty-third power.

And again it doesn’t matter if we make it the tenth root of 𝑥
all to the power of forty-three or the tenth root of 𝑥 to the power of
forty-three. Okay, hopefully that’s given you some good exposure to some quite tricky questions.
Rearranging complicated expressions in terms of a series of terms in power form, as I said
before, gonna be very helpful when it comes to differentiation and integration. Okay.