Question

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.

Answer 1

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

Answer 2

t= 7508.4 i think
I dont know how to do the second part

Answer 3

clearly <u>t= 7508.4 i think</u> 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

Answer 4

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