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
N(t) = 10,000/(5+1245e^-.97t)
How long before 1,000 people have heard it?

Answer 1

set that sucker equal to 1000 and solve for \(t\)

Answer 2

I've gotten to 10 = 5+1245e^-.97t
But don't know what to do next. Do I use natural log now?

Answer 3

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

Answer 4

subtract 5 from both sides

Answer 5

ok lets see what the first steps are
before getting to the log part

Answer 6

your goal is to write \[e^{-.97t}=\text{some number}\] Then use the log to solve for \(t\)

Answer 7

So I'll have to divide 5 by 1245?

Answer 8

so treat it like \[ 10 = 5+1245x\] sovle for \(x\)

Answer 9

yes

Answer 10

course if it was me, i would use technology, but you can do that on a test i guess

Answer 11

did you get to \[e^{-.97t}=\frac{1}{249}\] yet ?

Answer 12

I'm using my calculator. I have \[\frac{ 1 }{ 249 }\] =\[e ^{-.97t}\]

Answer 13

yup
two more steps

Answer 14

Now use natural log?

Answer 15

yes now

Answer 16

You can do another step to make writing easier :-)
\[249 = e^{0.97t}\]

Answer 17

I got -5.52. Would I set that equal to natural log -.97t? Then divide by -.97?

Answer 18

I divided -5.52 by -.97 and got 5.7. So t = 5.7?

Answer 19

Yes, t = 5.6881 is what i got.