anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
set that sucker equal to 1000 and solve for \(t\)
anonymous
  • anonymous
I've gotten to 10 = 5+1245e^-.97t But don't know what to do next. Do I use natural log now?
anonymous
  • anonymous
\[\frac{ 10,000}{5+1245e^{-.97t}}=1000 \]

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More answers

anonymous
  • anonymous
subtract 5 from both sides
anonymous
  • anonymous
ok lets see what the first steps are before getting to the log part
anonymous
  • anonymous
your goal is to write \[e^{-.97t}=\text{some number}\] Then use the log to solve for \(t\)
anonymous
  • anonymous
So I'll have to divide 5 by 1245?
anonymous
  • anonymous
so treat it like \[ 10 = 5+1245x\] sovle for \(x\)
anonymous
  • anonymous
yes
anonymous
  • anonymous
course if it was me, i would use technology, but you can do that on a test i guess
anonymous
  • anonymous
did you get to \[e^{-.97t}=\frac{1}{249}\] yet ?
anonymous
  • anonymous
I'm using my calculator. I have \[\frac{ 1 }{ 249 }\] =\[e ^{-.97t}\]
anonymous
  • anonymous
yup two more steps
anonymous
  • anonymous
Now use natural log?
anonymous
  • anonymous
yes now
whpalmer4
  • whpalmer4
You can do another step to make writing easier :-) \[249 = e^{0.97t}\]
anonymous
  • anonymous
I got -5.52. Would I set that equal to natural log -.97t? Then divide by -.97?
anonymous
  • anonymous
I divided -5.52 by -.97 and got 5.7. So t = 5.7?
whpalmer4
  • whpalmer4
Yes, t = 5.6881 is what i got.

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