# Simplifying Exponents in Fractions

From my Algebra 2 class. Not homework.

$$4x^3/2x^5y^2$$

Divide the bases and subtract the exponents:

$$2x^{-2}y^2$$

Get rid of negative exponent by division:

$$2y^2/x^2$$

$$2x^2y^2$$

Is this correct?

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assuming you mean $\frac{4x^3y^2}{2x^5}$ then you your first step is valid as is your second but note that "getting rid of the negative exponent by division" what you are really doing here is just writing out explicity what $x^{-2}$ represents. The final step however is not correct as $\frac{2y^2}{x^2} \neq 2x^2y^2$ –  Deven Ware Sep 15 '11 at 16:20
It's what I was told to do. I figured it was wrong anyway –  david Sep 15 '11 at 16:22
@david: If the original fraction is $\Bigl((4x^3)/(2x^5)\bigr)y^2$, then $(2y^2)/(x^2)$ is correct. But if the original fraction was $$\frac{4x^3}{2x^5y^2},$$ then the correct final answer should be $2x^{-2}y^{-2}$. –  Arturo Magidin Sep 15 '11 at 16:51
Note that fraction slashes often lead to ambiguity, as here. Please either use parentheses or stack the fraction vertically using \frac{numerator}{denominator} –  Ross Millikan Sep 15 '11 at 16:57

Here's how I think about the exponent rules. It may be helpful to you. Reduce the numerical part first and consider $$\frac{2x^3}{x^5y^2}.$$ I notice that both have common factors of $x$. How many copies of $x$ are there? There are 3 on the top and 5 on the bottom, so I could think of the fraction as $$\frac{2xxx}{xxxxxy^2}.$$ (Of course I would never actually write that down, but it's useful to keep in mind.) Now, if I went by and canceled factors of $x$ one-by-one, I would eventually be left with $$\frac{2}{xxy^2},$$ or better yet $$\frac{2}{x^2y^2}.$$ The advantage of thinking in this way is I don't memorize unmotivated rules about when to add or subtract exponents (though, if you reflect for a moment, you'll see that this method is the same as the exponent rules you've learned) and avoids getting negative exponents unnecessarily.

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This answer assumes that you intended the original problem to be $$\frac{4x^3}{2x^5y^2}$$ and not $$\frac{4x^3}{2x^5} \cdot y^2.$$ If you meant the latter, then it's all the same business except that $y^2$ will now be in the numerator. –  Austin Mohr Sep 15 '11 at 17:08

One of the problems with the "slant bar" notation is that it is not clear what fraction you have to begin with.

Does "$2x^2/x^5y^2$" represent $$\frac{2x^2}{x^5y^2},$$ or does it represent $$\frac{2x^2}{x^5}\cdot y^2\quad?$$

Normally, it would be interpreted as the first; but you seem to be interpreting it as the second. That would happen if, in the board or handwriting, there was a prominent and clear space, something like $$2x^2/x^5\ \ y^2,$$ which got lost along the way.

Nonetheless, because of this possibility of confusion, I strongly advice all my students to abandon the "slant bar" notation in mathematical formulas.

If your original problem was $$\frac{2x^2}{x^5y^2}$$ then your first step is wrong, because the $y^2$ is in the denominator. The final answer should be as Austin Mohr gives it.

If your original problem was $$\frac{2x^2}{x^5}\cdot y^2,$$ then everything you did was right up to the final step; you should not have gone from $\displaystyle\frac{2y^2}{x^2}$ to $\displaystyle 2y^2x^2$.

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