Question

6561^(2n)=729^(n+2) ummm N?

Answer 1

U know logarithm?

Answer 2

um no..

Answer 3

\[9^{4{\times}2n}=9^{4{\times}(n+2)}\]\[4\times(2n)=4\times(n+1)\]\[8n=4n+4\]\[4n=4\]\[n=1\]

Answer 4

so my n would equal 2?

Answer 5

you are asking me that " Is the sky green?"
I just say n=1

Answer 6

well. that was rude. 1 isnt one of my options

Answer 7

oh sorry it should be 4*(2n)=3*(n+1) w/c gives n = 3/5

Answer 8

did you get it?

Answer 9

not really but im like math awkward

Answer 10

ok 6561 = 9^4
729 = 9^3
Can u insert them in there respective place?

Answer 11

not really, tis why im here and tis why i failed this class last year.

Answer 12

guess you're gonna fail again ahahhhhhhhhhhhhhaahahaha

Answer 13

6561^(2n)=729^(n+2)

log (6561^2n) = log(729^[n+2])
2nlog(6561) = (n+2)log(729)

finish it.

Answer 14

that was so mean im a little butt hurt right now.

Answer 15

2nlog(6561) = nlog(729) + 2log(729)
2nlog (6561) - nlog (729) = 2log (729)
n(2log[6561] - log [729]) = 2log (729)
n = 2log (729) / [2log(6561) - log (729)]

There you go honey.

Answer 16

okay thanks boo.

Answer 17

remember, i don't like pickles on my big mac.

Answer 18

oh i thought you said you like extra pickles.. damn

Answer 19

hahahaha, so after you take a base of 27 from each logarithm, your final answer will be 6/5

Answer 20

clean up on isle 4 please?

Answer 21

6/5? oh goodness. im going to fail this is going to suck. thank you

Answer 22

hold up.

Answer 23

you spilt something in isle 4

Answer 24

oh heeeellllll no

Answer 25

what do you not understand about my steps?

Answer 26

i simply took the logarithm base 27 from both sides, applied the power rule, factored and isolated for n.

Answer 27

no, its not you really. its me. i'm math challenged. i question stuff too much

Answer 28

wait that kinda made sense!

Answer 29

oh who's a big boy now?