Here's the question you clicked on:
Just curious. It marked it wrong but I'm sure I got it right. Look: x^2 ---- x^6 = 1 ---- x^-4 Now to make it positive we put it in the numerator x^4 ---- 1 Which is just equal to x^4
\[\large \frac{x^2}{x^6} \qquad = \qquad x^{2-6} \qquad = \qquad x^{-4}\]Do you see the mistake you made? You had the correct power, but when you divide, you should be left with your X in the numerator, not the bottom one.
I don't think so. \[\frac{ x^2 }{ x^6 } \rightarrow \frac{ 1 }{ x^6 - x^2 } \rightarrow \frac{ 1 }{ x^-4 } \rightarrow \frac{ x^4 }{ 1 }\rightarrow x^4\] We subtract the denominator from the numerator.
\[\large \dfrac{x^a}{x^b} = x^{a - b} = \dfrac{1}{x^{b - a}}\]
You could write it like this if you wanted.\[\large \frac{x^2}{x^6} \qquad = \qquad \frac{1}{x^6x^{-2}} \qquad =\qquad \frac{1}{x^{6-2}} \qquad =\qquad \frac{1}{x^4} \qquad = \qquad x^{-4}\]
Zepdrix is right, you know.
You have the right idea (although the notation is a bit sloppy). You just made a tiny mistake, when you subtracted 2 from 6, it should give you `positive` 4.
If that's true, it should be -4, \[\frac{ 1 }{ x^4 }\] would be fine.
Yes that would be a fine answer! :)
Yeah. Most people hate negative exponents, so it's good enough.
But if... \[\frac{ x^6 }{ x^5 } = 6 - 5\] Why would \[\frac{ x^5 }{ x^6} = 6 - 5, too?\]
No, look at my 'formula' above here.
\[\large \frac{x^5}{x^6}=x^{5-6}\] Yah your numbers are a little backwards on the second example. :) You always subtract the BOTTOM number, it doesn't matter which number is smaller. Always subtract the power in the denominator.
lololol good ole morgan freeman XDDD