# Division with negative exponents

I have a problem that looks like this:

$$\frac{20x^5y^3}{5x^2y^{-4}}$$

Now they said that the "rule" is that when dividing exponents, you bring them on top as a negative like this:

$$4x^{5-2}*y^{3-(-4)}$$

That doesn't make too much sense though. A term like $y^{-4}$ is essentially saying $\large \frac 1{y^4}$ in the denominator because a negative exponent is the opposite of a positive exponent and you use division. And so here you are dividing by $y$ four times. So if that's the case, you cross multiply: $\large \frac{1}{y^4} \frac{y^4}{1}$ on bottom and then of course to keep balance, you multiply $\large \frac{y^4}{1}$ on top to get this:

$$4x^{5 - 2}y^{3 + 4}$$

Now look at my solution and look at the other one. They get the same answer but through different means. I dont see how they get $y^{3-(-4)}$.

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$\frac{y^p}{y^q} = y^{p-q}$ for any $p$ and $q$, positive or negative. You just need to know how to subtract negative numbers. I think you are overthinking this. –  Stefan Smith Apr 16 '13 at 23:52

What the other solution did is a straightforward application of the rule: $\frac1{y^n} = y^{-n}$. For $n = -4$, you get $\frac1{y^{-4}} = y^{-(-4)} = y^4$. Does that make it clear?

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There is nothing wrong with your thinking. You are correct.

But so is the text: Note that you do indeed arrive at the same answer.

$$4x^{5-2}y^{3-(-4)}= 4x^{5-2}y^{3+4} = 4x^3y^7$$

The textbook did exactly what you both did with $\dfrac{x^5}{x^2} = x^{5-2}$: subtracting the exponent $2$ which is in the denominator, from the exponent of $x$ in the numerator: $x^{5-2}$ Since $y^{-4}$ is in the denominator, we subtract $-4$ from the exponent of $y$ in the numerator, giving $y^{3-(-4)} =y^{3+4}$.

Note that what the text refers to as a "rule" is simply a method for obtaining the correct value of the exponent. But the rule is based in the laws of exponents, and can be similarly justified consistently with the logic you applied.

You approached this, if I understand correctly, like this:

$$\frac{20x^5y^3}{5x^2y^{-4}} = 4x^{5-2}\cdot \frac{y^3}{\frac 1{y^{4}}} \cdot \frac{y^4}{y^4} = 4x^{5-2}y^{3+4} = 4x^3 y^7$$

And that's a perfectly fine way to handle the problem.

Essentially, what's happening here is that we have $$\frac{20x^5y^3}{5x^2y^{-4}} = \dfrac{\not{5}\cdot 4 \cdot \not{x^2}\cdot x^3\cdot y^3\cdot y^{4}}{\not{5}\cdot \not{x^2}\cdot {\large \frac{1}{{ \not{y^4}}}}\cdot \not{y^4}} = 4x^3y^{3+4} = 4x^3y^7$$

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