Question

Simplify the following expression, and rewrite it in an equivalent form with positive exponents.
(9x-3)-3

Answer 1

hint: \[\huge a^{-m} \implies \frac{1}{a^m}\]
so what do you think is the value of \[\huge (9x- 3)^{-3}\]

Answer 2

1/(9x-3)^3

Answer 3

right

Answer 4

problem solved?

Answer 5

so its simplified?? thats it thats all???

Answer 6

i believe so...unless you'd like to expand that denominator you ahve

Answer 7

but guess what?? I just looked at the problem again, it didnt copy over right. it should be

(9x^3)^3
does it still apply

Answer 8

well kinda...

Answer 9

is that \[(9x^3)^3\] or \[(9x^3)^{-3}\]

Answer 10

the x has a negative 3 too

Answer 11

so it's \[\huge (9x^{-3})^{-3}?\]

Answer 12

yes thats it, so will i still use the same equation you gave at first

Answer 13

yep \[\huge a^{-m} = \frac {1}{a^m}\]

Answer 14

so what do you have?

Answer 15

yeah i did 1/(9x^3)^3

Answer 16

dont you mean \[\huge \frac{1}{(9x^{-3})^3}\]

Answer 17

ohhhhh so I am only changing the outer exponent??

Answer 18

so far yes.

Answer 19

now use the rule \[\huge (a^b x^y)^c \implies a^{bc} x^{yc}\]

Answer 20

wow okay 729x^-27 ?????

Answer 21

right

Answer 22

now use \[\huge a^{-m} \implies \frac{1}{a^m}\] on x^-27

Answer 23

1/x^27

Answer 24

right...so all in all that would be..?

Answer 25

here's a hint: 2x^-3 would be 2/x^3

so what would be 729 x^-27

Answer 26

729x^27

Answer 27

nope...remember what a^-m is

Answer 28

like i said \[\huge 2x^{-3} \implies \frac{2}{x^3}\]
so what's \(729 x^{-27}\)

Answer 29

1/729x^27