# precalculas

posted by on .

the amount of bacteria in a petri dish increased by a percent change of 12% each hour over a period of 15 hours.

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 - ,

a. 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 - ,

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 - ,

I 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 - ,

drwls 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 ?)