Question 1144616: Please help me solve this problem. "A researcher is investigating a specimen of bacteria. She finds that the original 1,000 bacteria grew to 2,084,000 in 60 hours. How fast does the bacteria (a) double? (b) quadruple?". Please kindly explain to me how. Thank you in advance.
Found 2 solutions by Alan3354, rothauserc:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! "A researcher is investigating a specimen of bacteria. She finds that the original 1,000 bacteria grew to 2,084,000 in 60 hours. How fast does the bacteria (a) double? (b) quadruple?".
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Confirm it's 2084000 and not 2048000
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If it's 2048 that's 2^11.
60/12 = 5 hour periods
In 1000's and periods of 5 hours:
0 1
5 2
10 4
15 8
20 16
25 32
30 64
35 128
40 256
45 512
50 1024
55 2048
================
Using 60/11 hours for the periods:
P #
===========
0 1
1 2
2 4
3 8
4 16
5 32
6 64
7 128
8 256
9 512
10 1024
11 2048
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! The exponential/decay formula is
:
A = Pe^(rt), A is the ending amount(for example: bacteria, or radioactive element), P is the beginning amount, r is the growth or decay rate, t is time, e is a constant
:
For this problem, the units of t are hours(t = 60), at t = 0, P = 1,000
:
First, we want to find the value for r
:
2084000 = 1000 * e^(r * 60)
:
2084 = e^(r * 60)
:
take natural logarithm of both sides of =
:
60r = ln(2084)
:
r = ln(2084)/60 = 0.1274/hour
:
(a) 2000 = 1000 * e^(0.1274 * t)
:
2 = e^(0.1274t)
:
0.1274t = ln(2)
:
t = ln(2)/0.1274 = 5.44 hours
:
(b) 4000 = 1000 * e^(0.1274 * t)
:
4 = e^(0.1274 * t)
:
t = ln(4)/0.1274 = 10.88 hours
:
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