Exponent properties 2

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.