Question

(a^5b^-7)(a^-4b^9)

trying to check my answer

Answer 1

do you multiply the exponents? and leave it at a^-20b^-63

Answer 2

No you add the corresponding exponents

Answer 3

\[a^xa^y=a^{x+y}\]

Answer 4

ex:

a^5 times a^-4 = a^(5 + (-4) ) = a^(5 - 4) = a^1 = a


So a^5 times a^-4 = a

Answer 5

so thatll be ab^2

Answer 6

you got it

Answer 7

thanks, are you good with dividing these?

Answer 8

Yes

Answer 9

dividing is the opposite: you subtract corresponding exponents

Answer 10

(2a^-3b^2)^-1
---------
(6a^2b^4)

Answer 11

The first thing to do is to simplify (2a^-3b^2)^-1

Do you know how to do that?

Answer 12

add -1 to the exponents?

Answer 13

2a^-4b?

Answer 14

No you first write 2 as 2^1


Then you multiply the outer exponent of -1 by every exponent


So


(2a^-3b^2)^-1


(2^1a^-3b^2)^-1


2^(1*-1) * a^(-3*-1) * b^(2*-1)


2^(-1) a^3 b^(-2)


This means that (2a^-3b^2)^-1 simplifies to 2^(-1) a^3 b^(-2)

Answer 15

it looks really confusing on the computer screen with all the ^'s

Answer 16

wait a second, was that -1 outside the entire group or just outside the numerator?

Answer 17

entire group

Answer 18

is the original problem


\[\Large \left(\frac{2a^{-3}b^{2}}{6a^2b^4}\right)^{-1}\]


OR...


is it


\[\Large \frac{\left(2a^{-3}b^{2}\right)^{-1}}{6a^2b^4}\]

Answer 19

first one

Answer 20

alright thx for clarifying, one sec

Answer 21

sorry i shouldve wrote it a little better

Answer 22

no worries

Answer 23

Here's how you tackle this problem...


\[\Large \left(\frac{2a^{-3}b^{2}}{6a^2b^4}\right)^{-1}\]


\[\Large \left(\left(\frac{2}{6}\right)\left(\frac{a^{-3}}{a^2}\right)\left(\frac{b^{2}}{b^4}\right)\right)^{-1}\]



\[\Large \left(\left(\frac{1}{3}\right)\left(\frac{a^{-3}}{a^2}\right)\left(\frac{b^{2}}{b^4}\right)\right)^{-1}\]


\[\Large \left(\left(\frac{1}{3}\right)a^{-3-2}*b^{2-4}\right)^{-1}\]


\[\Large \left(\left(\frac{1}{3}\right)a^{-5}*b^{-2}\right)^{-1}\]


\[\Large \left(\left(\frac{1}{3}\right)\left(\frac{1}{a^5}\right)\left(\frac{1}{b^2}\right)\right)^{-1}\]


\[\Large \left(\frac{1}{3a^5b^2}\right)^{-1}\]


\[\Large \left(\frac{3a^5b^2}{1}\right)^{1}\]


\[\Large 3a^5b^2\]


So \[\Large \left(\frac{2a^{-3}b^{2}}{6a^2b^4}\right)^{-1}\] fully simplifies to \[\Large 3a^5b^2\]

Answer 24

The idea in the second to last step is that when you have a fraction raised to a negative exponent, you flip the fraction to make the exponent positive.

Answer 25

thanks bro helped alot!!

Answer 26

you're welcome