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April 18, 2014

April 18, 2014

Posted by **katrina** on Tuesday, March 1, 2011 at 12:28am.

a. find the growth factor for 1 hour?

b. by what total percent did the bacteria change during this 15- hour time period.?

c. how long will it take for the bacteria to double?

- precalculus -
**drwls**, Tuesday, March 1, 2011 at 9:16ama. 1.12

b. [(1.12)^15 -1] x 100

The x 100 is for conversion %

c. about 6 hours, using the "rule of 72"

For the exact answer (t), solve

(1.12)^t = 2

- precalculas -
**Reiny**, Tuesday, March 1, 2011 at 9:18amyour formula would be

Amount = a(1.12)^t , where 0 ≤ t ≤ 15

a) in 1 hour growth factor would be 1.12

b) Amount = a(1.12)^15 = 5.47a

percentage of growth = 5.47a/a = 5.47 or 547%

c)

2a = a(1.12)^t

2 = 1.12^t

log2 = log(1.12^t)

log2 = t log1.12

t = log2/log1.12 = 6.1

- precalculas -
**drwls**, Tuesday, March 1, 2011 at 9:25amI do not agree with Reiny's answer #2. They asked for the change, so 1 must be subtracted from 1.12^15

This would result in a 447% increase, to 547% of the initial value

- precalculas -
**Reiny**, Tuesday, March 1, 2011 at 9:34amdrwls is correct, I should have subtracted to get the change.

(Can I blame it on the fact that I was only on my first cup of coffee ?)

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