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simplify three A to the fifth over nine A squared times A to the fourth over A to the third

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so before we even worry about the A's we can actually simplify the three and the nine

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they're both divisible by three

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so let's divide the numerator and the denominator here by three

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so if we divide the numerator by three, the three becomes a one

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if we divide the denominator by three, the nine becomes a three

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so this reduces to, or simplifies to, one A to the fifth times A to the fourth

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or maybe i should say A to the fifth over three A squared

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times A to the fourth over A to the third

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now this, if we just multiply the two expressions this would be equal to

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one A to the fifth times A to the fourth in the numerator

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and we don't have to worry about the one, it doesn't change the values

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so it's A to the fifth times A to the fourth in the numerator

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and then we have three times A squared times A to the third in the denominator

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now there's multiple ways that we can simplify this from here

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one, sometimes its called the quotient rule, and its just the idea that if you have

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A to the 'x' over A to the 'y', that this is going to be equal to A to the 'x' minus 'y'

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and just to understand why that works lets think about A to the fifth over A squared

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so A to the fifth is literally A times A times A times A times A

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that right there is A to the fifth

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and we have that over A squared, and i'm just thinking about the A squared right over here

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so we have that over A squared, which is literally just A times A

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that is A squared

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now we can clearly, both the numerator and the denominator, are both divisible by A times A

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we can divide them both by A times A

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if we divide the numerator by A twice, so by A times A, so lets get rid of an A times A

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and if we divide the denominator by A times A, we just get a one

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so what are we just left with

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we are left with just A times A times A over 1, which is just A times A times A

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but what is this?

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this is A to the third power, or A to the five minus two power

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we had five, we were able to cancel out two, that gave us three

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so we can do the same thing over here, we can apply the quotient rule

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and i'll do this in two ways, so lets apply the quotient rule with the A to the fifth and the A squared

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and also, let me do it this way

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so lets apply with these two guys and then lets apply it with these two guys

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and of course we have the one third out front

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so this can be reduced to one third times, if we apply the quotient rule with A to the fifth over A squared

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we just did it over here

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that becomes A to the third power

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and if we apply it over here with the A to the fourth over A to the third

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that'll give us A to the four minus three power

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or A to the first power

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and of course we can simplify this as A to the third times A

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well actually let me just do it over here

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before I even rewrite it we know that A to the third times A to the first

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is going to be A to the three plus one power

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we have the same base so we can add the exponents

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we're multiplying A times itself three times and then one more time

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so that'll be A to the fourth power

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so this right over here becomes A to the fourth power, A to the three plus one power

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and then we have to multiply that by one third

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so our answer could be one third A to the fourth

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or we could equally write it A to the fourth over three

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now the other way to do this problem would've been to apply the product

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or to add the exponents in the numerator and then add the exponents in the denominator

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so lets do it that way first

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so if we add the exponents in the numerator first

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we don't apply the quotient rule first, we apply it second

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we get in the numerator A to the fifth times A to the fourth

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would be A to the ninth power, five plus four

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and then in the denominator we have A squared times A to the third

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add the exponents because we were taking the product with the same base

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so it'll be A to the fifth power

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and we still have the three down here, we have a one third, or we could just write a three over here

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now we could apply the quotient property of exponents

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we could say look we have A to the ninth over A to the fifth

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A to the ninth over A to the fifth is equal to A to the nine minus five power

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or it's equal to A to the fourth power

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and of course we still have the divided by three

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either way we got the same answer