Question

Division properties of exponents...

Answer 1

which of the following is equivalent to \[\frac{ x^{5}y^{2} }{ xy^2}\] when \[x \neq 0\] and \[y \neq 0\]

Answer 2

can someone help me learn how to do this?

Answer 3

when you divide, you subtract the exponents
it is really the same as cancelling

Answer 4

for example \[\frac{x^5}{x}=x^{5-1}=x^4\]

Answer 5

Yes, and to simplify matters a bit, concentrate on x first; then concentrate on y, separately:

\[\frac{ x^5 }{ x }*\frac{ y^2 }{ y^2}\]

Answer 6

that is because \[\frac{x^5}{x}=\frac{x\times x\times x\times x\times x}{x}\\
=\frac{x\times x\times x\times x\times\cancel{x }}{\cancel{x}}=x^4\]

Answer 7

then \[\frac{y^2}{y^2}=1\] since anything divided by itself is \(1\)

Answer 8

I think I understand this now, thank you, and does the 1 get written anywhere? or is it just \[x^{4}\]

Answer 9

it is just \(x^4\)

Answer 10

okay, thank you so much (:

Answer 11

yw

Answer 12

@satellite73 what happens to the 1 does it just disappear? Im curious

Answer 13

or @mathmale

Answer 14

@TheSmartOne can you explain?

Answer 15

sure, one sec

Answer 16

which of the following is equivalent to \[\frac{ x^{5}y^{2} }{ xy^2}\] when \[x \neq 0\] and \[y \neq 0\]

~~~~~~~~~~~~~~~~~~

\(\LARGE\frac{ x^{5}y^{2} }{ xy^2}\)

Ok, now we can break up this fraction, since all we're doing is multiplying. If we were adding the terms, we wouldn't be able to do what I'm about to do:

\(\LARGE \frac{x^5 y^2}{xy^2} = \frac{x^5}{x} \times \frac{y^2}{y^2}\)

And as satellite told you, \(\sf\Large \frac{y^2}{y^2} = y^{2-2} = y^0 = \boxed{1}\)

Now, when we simplify what we had, we get:

\(\sf\LARGE \frac{x^5}{x} \times \frac{y^2}{y^2} = \frac{x^5}{x} \times 1 = \boxed{\frac{x^5}{x}}\)

because as you know, anything times 1 is one.

and then from there you know that

\(\sf\Large \frac{x^5}{x} = x^{5-1} = \boxed{x^4}\) which was your answer :)

Answer 17

if you have anymore questions, just ask :)

Answer 18

I see now, thank you! I didn't see that you multiply all of them together that way at first

Answer 19

I'm glad that you understand it better! :)