Question

The number of people in a town of 10,000 who have heard a rumor started by a small group of people is given by the following function : N(t) = 10,000 / 5 + 1245e^-0.97t
How long will it be until 1,000 people in the town have heard the rumor?

Answer 1

oh this one again

Answer 2

you want a long method, or go straight to the answer?

Answer 3

i had the decimal in the wrong place, is this i think\[\huge N(t)=\frac{10,000}{5+1245e^{-0.97t}}\]

Answer 4

Yeah, I solved it but the answer seemed wrong. The answer would be fine and I can just use what I know to show the work!

Answer 5

you set it equal to 1000 and start with
\[\frac{10,000}{5+1245e^{-0.97t}}=1000\]

Answer 6

That's what I did, but I got a decimal for my answer although my math could be wrong ha

Answer 7

then some algebra
\[10,000=1000(5+1245e^{-.97t})\\
10=5+e^{-.97t}\\
5=e^{-.97t}\]

Answer 8

the in log form
\[\ln(5)=-.97t\] so
\[t=\frac{\ln(5)}{-.97}\]

Answer 9

ok i made a mistake somewhere

Answer 10

ooho i see it

Answer 11

\[10,000=1000(5+1245e^{-.97t})\\
10=5+1245e^{-.97t}\\
5=1245e^{-.97t}\]

Answer 12

so
\[\frac{5}{1245}=e^{-.97t}\] and then \[\frac{1}{249}=e^{-.79t}\]
\[\ln(\frac{1}{249}=-.97t\\
t=\frac{\ln(1/249)}{-.97}\]

Answer 13

i get about \(5.7\)

Answer 14

http://www.wolframalpha.com/input/?i=-log%281%2F249%29%2F.97

Answer 15

@satellite73 Thank you so much! My computer was acting up so I had to restart it. But I really appreciate your help!