Question

I'm stuck at a simplifying-indices problem...

[9^2 x 3^(n-1)]/[3^n + 3^(n-1)]

Answer 1

http://mathway.com/answer.aspx?p=calg?p=(9SMB072SMB15xSMB073(n-1)0)SMB10(3SMB07nSMB15+SMB153SMB07(n-1))?p=7?p=?p=?p=?p=?p=0?p=?p=0?p=?p=?p=Simplify

Answer 2

The answer key on the book says 27/4, though. D: There's this rule that I don't understand.
Oh, and the "x" is multiplication, btw... just noticed that.

Answer 3

-_-

for multiplication people usually use * for a multiplication not x

Answer 4

There's no way you end up with 27/4 if that's your original problem.

Answer 5

I get 81/4 btw

Answer 6

@BVB_army, haha, sorry. Zombie-typing, I guess. :P
@Hero, ... Really? How'd you get that? What I don't understand is how to take the common factor from the denominator. D:

Answer 7

All you have to do is utilize rules of exponents to simplify

Answer 8

\[3^{n-1}=3^n \times 3^{-1}\]

Answer 9

... Uh, so then it becomes. 3^n + 3^n/3. Right?
Then it should become 4(3^(n-1)), right? <-- That's the part I don't get... sorry to bother you with all these questions. D: Thank you, thank you.

Answer 10

The steps for this is quite interesting. I'll have to show you on vyew.

Answer 11

I'm not exactly sure what a Vyew is... but I'm making the account now.

Answer 12

@Hero, thank you again. :D

Answer 13

yw