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.