Question

Pre-calc help please?

During a research experiment, it was found that the number of bacteria in a culture grew at a rate proportional to its size. At 5:00 AM there were 5,000 bacteria present in the culture. At noon, the number of bacteria grew to 5,600. How many bacteria will there be at midnight?

Answer 1

\[5000e^7=5600 --> 7k= \ln(5600/5000)> t(14)=5000e^.01618981\]
@Photon336 this is as far as I got can you help me please?

Answer 2

@ILovePuppiesLol can you help me with this please?

Answer 3

@jim_thompson5910 can you help me with this please?

Answer 4

so they gave you a formula or did they want you to derive one out?

Answer 5

they wanted me to derive one

Answer 6

I'm guessing you haven't learned about derivatives?

Answer 7

not really no

Answer 8

hmm let me think

Answer 9

did your teacher give you a template to work with?

Answer 10

A= P(1+r/n)^n/t
and A= Pe^rt

Answer 11

ok it sounds like we'll be using A= Pe^(rt)

Answer 12

A= Pe^(rt)

A= Pe^(r*0) ... plug in t = 0 for time 0

5000 = P*e^(r*0) .... plug in the population at time 0

solve for P to get P = 5000

agreed so far?

Answer 13

yep

Answer 14

ok so

A = 5000*e^(r*t)

is the equation so far

Answer 15

yea

Answer 16

now plug in A = 5600 and t = 7 (noon is seven hours after 5 AM)

A = 5000*e^(r*t)
5600 = 5000*e^(r*7)

then solve for r

Answer 17

it looks like you were doing that at the top now that I look it over more thoroughly

Answer 18

oh ok, well i got .01618981 for r

Answer 19

me too

Answer 20

the equation is now

A = 5000*e^(0.01618981*t)

Answer 21

plug in t = 19 (add 12 more hrs to t = 7 to get t = 19 to get to midnight)

A = 5000*e^(0.01618981*t)
A = 5000*e^(0.01618981*19)
A = ??

Answer 22

i got 6800.827

Answer 23

so roughly 6801 bacteria

Answer 24

ok thanks so much!

Answer 25

you're welcome