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 100 bacteria initially, and 500 bacteria 5 hours later.

or 201 growing s hours lter the rate of growth, k, of the btria Loe or erms of logarithms that can model the growth of the hacteria at time, Ltave your answer in terms of logarithms #10. Round your answer to the nearest whole bacteria. Show all steps using the properties of logarithms that lead to your answer. C. Determine the number of bacteria that will be present at time t the work that leads to your answer and round the answer to the nearest hour. . Determine the time at which the number of bacteria reaches 12,500. Show

Answer 1

Solution:

(A) The given equation is

$y=ae^{kt}$ ------------------------------------------------ (1)

at t=0 (initially), y=100 bacteria. So, from eqn 1, substitute t=0 and y=100. So,

$100=ae^{k*0}=ae^0=a$ ------------------------------ (2)

at t=5 hrs, y=500 bacteria. So, using eqn 1 and substituting t=5, y=500,

$500=ae^{k*5}=ae^{5k}$ ----------------------------------- (3)

Now, substituting a=100 from eqn 2 into eqn 3.

$500=100e^{5k}$

dividing both sides by 100

$\frac{500}{100}=\frac{100e^{5k}}{100}$  

$5=e^{5k}$

taking natural logarithm on both sides

$ln(5)=ln(e^{5k})$

$ln(5)=5k$

dividing both sides by 5

$\frac{ln(5)}{5}=\frac{5k}{5}$  

$\frac{ln(5)}{5}=k$ --------------------------------------------- (4)

So, the rate of growth, $k=\frac{ln(5)}{5}$

(B) Substituting eqn 2 and 4 into eqn 1

$y=100e^{\frac{ln(5)}{5}t}$ ----------------------------- (5)

This is the equation for modeling the growth of bacteria at time, t.

(C)  at time, t=10 hrs. Using eqn 5

$y=100e^{\frac{ln(5)}{5}\times 10}$

$y=100e^{2(ln5)}$

$y=100e^{(ln5^2)}=100e^{ln25}$ [since, aln(b)=ln(b^a)]

$y=100*25$ [since, e^(lnx)=x. So, e^(ln25)=25]

$y=2500$

So, after t=10 hrs. there will be 2500 bacteria.

(D) t=?, y=12500. So

$12500=100e^{\frac{ln5}{5}t}$

dividing both sides by 100

$\frac{12500}{100}=\frac{100e^{\frac{ln5}{5}t}}{100}$

$125=e^{\frac{ln5}{5}t}$

taking logarithm on both sides

$ln(125)=\frac{ln5}{5}t$

$t=5\times \frac{ln(125)}{ln5}$

$t=5\times \frac{ln(5^3)}{ln5}$

$t=5\times \frac{3ln(5)}{ln5}$

$t=5\times 3$

$t=15$ hrs.

So, after t=15 hrs, there will ne 12500 bacteria.