find k using
300 e^(5k) = 1500
use that k with t=10
finally, use that k to find t when
300 e^(kt) = 12500
A certain type of bacteria is growing at an exponential rate that can be modeled by the equation y=ae^kt, where t represents the number of hours. There are 300 bacteria initially, and 1500 bacteria 5 hours later.
How many bacteria will be present at time t=10? Find you answer to the nearest whole bacteria. Show all steps using properties of logarithms
Determine the time at which the number of bacteria reaches 12,500. Round your answer to the nearest hour.
4 answers
t= 7508.4 i think
I dont know how to do the second part
I dont know how to do the second part
clearly t= 7508.4 i think is not a reasonable answer. 7500 years?
300 e^(5k) = 1500
e^(5k) = 5
5k = ln5
k = ln5/5 = 0.32
so,
y(t) = 300 e^(0.32t)
y(10) = 300 e^3.2 = 7360
300 e^(0.32t) = 12500
e^(0.32t) = 41.67
0.32t = ln 41.67 = 3.73
t = 3.73/0.32 = 11.65
300 e^(5k) = 1500
e^(5k) = 5
5k = ln5
k = ln5/5 = 0.32
so,
y(t) = 300 e^(0.32t)
y(10) = 300 e^3.2 = 7360
300 e^(0.32t) = 12500
e^(0.32t) = 41.67
0.32t = ln 41.67 = 3.73
t = 3.73/0.32 = 11.65
Y=ae^-kt
1500=300e^k4
ln5=e^k4 ln
1.609=-k4
*divide by 4 and by -1 to get rid of the negative
O.4024=k
1500=300e^k4
ln5=e^k4 ln
1.609=-k4
*divide by 4 and by -1 to get rid of the negative
O.4024=k
1 answer