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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 AIDS we can actually simplify the 3 and the 9 they're both divisible by 3 so let's divide the denominator and the 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 1 a to the fifth times a to the 4th over or maybe I should say a to the fifth over 3 a squared times a to the fourth over a to the 3rd now this if we just multiply the two expressions this would be equal to 1 a to the fifth times a to the 4th in the numerator and I we don't have to worry about the 1 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 3/8 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 1 sometimes it's called the quotient rule and that's just the idea it'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 why that works let's think about a to the fifth over a squared so a to the fifth is literally 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 so we have that over a squared which is literally just a times a that is a squared now we can clearly both the numerator and the 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 an 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 one which is just a times a times a a times a times a but what is this this is a to the third power or a to the 5-3 or sorry a to the 5-2 power we had five we were able to cancel out to that gave us three so we can 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 and also so let's let me do it this way so let's apply with these two guys and then let's apply it and then let's apply it with these two guys and of course we have the one-third we have the one-third out front so this can be reduced to one-third times if we apply the quotient rule with a to the fifth over eight squared we just applied 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 that'll give us let me do it that same blue color that'll give us a that's not the same blue color there we go this this will give us a to the four 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 over before we even rewrite it we know that a to the third times a to the first is going to be a to the three plus one 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 three plus one power and then we have to multiply that by one-third so our answer could be one-third a to the fourth or we could equally equally write it a to the fourth over three 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 so 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 five plus four and then in the denominator in the denominator we have a squared times a to the third add the exponent 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 three down here we have a one-third or we could just write a three over here now we could apply now we can 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 three we still have the divided by three either way either way we got the same answer