Simplify 3a to the fifth

over 9a squared times a to the fourth over

a to the third. So before we even

worry about the a’s, we can actually simplify

the 3 and the 9. They’re both divisible by 3. So let’s divide the numerator

and the denominator here by 3. So if we divide the numerator

by 3, the 3 becomes a 1. If we divide the denominator

by 3, the 9 becomes a 3. So this reduces

to, or simplifies to 1a to the fifth

times a to the fourth over– or maybe I should

say, a to the fifth over 3a squared times a to the

fourth over a to the third. Now this, if we just

multiply the two expressions, this would be equal to 1a to

the fifth times a to the fourth in the numerator, and we don’t

have to worry about the one, it doesn’t change the value. So it’s a to the fifth times a

to the fourth in the numerator. And then we have 3a– let

me write the 3 like this– and then we have 3

times a squared times a to the third in

the denominator. And now there’s

multiple ways that we can simplify this from here. One sometimes is called

the quotient rule. And that’s just the idea that

if you have a to the x over a to the y, that this is going to

be equal to a to the x minus y. And just to understand

why that works, let’s think about a to

the fifth over a squared. So a to the fifth

is literally a times a times a times a times a. That right there

is a to the fifth. And we have that over a squared. And I’m just thinking about

the a squared right over here, which is literally

just a times a. That is a squared. Now, clearly, both the

numerator and denominator are both divisible by a times a. We can divide them

both by a times a. So we can get rid of– if

we divide the numerator by a twice, by a times a, so

let’s get rid of a times a. And if we divide the denominator

by a times a, we just get a 1. So what are we just left with? We are left with just a

times a times a over 1, which is just a times a times a. But what is this? This is a to the third power,

or a to the 5 minus 2 power. We had 5, we were able to

cancel out 2, that gave us 3. So we could do the

same thing over here. We can apply the quotient rule. And I’ll do two ways

of actually doing this. So let’s apply the

quotient rule with the a to the fifth and the a squared. So let me do it this way. So let’s apply with

these two guys, and then let’s apply

it with these two guys. And of course, we have

the 1/3 out front. So this can be

reduced to 1/3 times– if we apply the

quotient rule with a to the fifth over

a squared, we just did it over here– that

becomes a to the third power. And if we apply it over here

with the a to the fourth over a to the third, that’ll

give us a– let me do it that same blue color. That’ll give us a– that’s

not the same blue color. There we go. This will give us a to

the 4 minus 3 power, or a to the first power. And of course, we can

simplify this as a to the third times

a– well, actually, let me just do it over here. Before I even rewrite it, we

know that a to the third times a to the first is going to

be a to the 3 plus 1 power. We have the same base,

we can add the exponents. We’re multiplying a times

itself three times and then one more time. So that’ll be a to

the fourth power. So this right over here

becomes a to the fourth power. a to the 3 plus 1 power. And then we have to

multiply that by 1/3. So our answer could be

1/3 a to the fourth, or we could equally write

it a to the fourth over 3. Now, the other way

to do this problem would have been to

apply the product, or to add the exponents

in the numerator, and then add the exponents

in the denominator. So let’s do it that way first. If we add the exponents

in the numerator first, we don’t apply the

quotient rule first. We apply it second. We get in the numerator, a to

the fifth times a to the fourth would be a to the ninth power. 5 plus 4. And then in the denominator

we have a squared times a to the third. Add the exponents, because

we’re taking the product with the same base. So it’ll be a to

the fifth power. And of course, we still

have this 3 down here. We have a 1/3, or we could

just write a 3 over here. Now, we could apply the

quotient property of exponents. We could say, look, we have a to

the ninth over a to the fifth. a to the ninth

over a to the fifth is equal to a to

the 9 minus 5 power, or it’s equal to a

to the fourth power. And of course, we still

have the divided by 3. Either way we got

the same answer.

thanks

@paulceltics no problem

Ur vids save my life

I love khan acedemy

sal this stuuf is stupid but like wheen u teach it its better

why do you add 1/3 at the front?

He is a blessing from G0d

how do you solve (1/2t^3)(2/3r^2)