A small population of fruit flies doubles in size every 10 days. If there are 20 fruit flies today, on what date will there be 100 fruit flies? Use the equation #P(t) = P_0e^(kt)# to solve this problem.

1 Answer
marfre
Apr 24, 2018

#~~23.22 # days

Explanation:

Given: #P(t) = P_oe^(kt); " "P_o = 20#; when #t = 10# days the number of fruit flies double. How many days will it be when there are #100# fruit flies.

First find #k# using #P_o = 20#; when #t = 10# days the number of fruit flies double:

Double of #20 = 2 *20 = 40# flies

#40= 20e^(k*10)#

#40/20 = 2 = e^(k*10)#

Use the logarithmic property: #ln e^x = x#

#ln 2 = ln e^(k*10)#

#ln 2 = k*10#

#k = (ln 2)/10#

Now find how long it will take for 100 fruit flies to exist:

#P(t) = P_oe^(kt)#

#100 = 20 e^((ln 2)/10 *t)#

#100/20 = e^((ln 2)/10 *t)#

#5 = e^((ln 2)/10 *t)#

#ln 5 = ln e^((ln 2)/10 *t)#

#ln 5 = (ln 2)/10 *t#

#10/(ln 2) * ln 5 = t#

#t = (10 ln 5)/(ln 2) ~~ 23.22# days

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