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 lets 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 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 if we divide the numerator by A twice, so by A times A, so lets get rid of an A times A and if we divide the denominator by A times A, we just get a one 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 five minus two power we had five, we were able to cancel out two, that gave us three so we can do the same thing over here, we can apply the quotient rule and i'll do this in two ways, so lets apply the quotient rule with the A to the fifth and the A squared and also, let me do it this way so lets apply with these two guys and then lets apply it with these two guys and of course 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 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 to the four minus three 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 three plus one power we have the same base so 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 write it A to the fourth over three now the other way to do this problem would've been to apply the product or to add the exponents in the numerator and then add the exponents in the denominator so lets 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 we have A squared times A to the third add the exponents because we were taking the product with the same base so it'll be A to the fifth power and we still have the three down here, we have a one third, or we could just write a three 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