Question

A population of bacteria is initially 6000. After three hours the population is 3000. If this rate of decay continues, find the exponential function that represents the size of the bacteria population after t hours. Write your answer in the form f(t)=a(b)t.

Answer 1

Answer:

The equation is $A=6000e^{-0.23t}$  at a rate of -23%.

Step-by-step explanation:

Decay can be represented by the equation$A=A_0e^{rt}.$ We can find the rate at which it decays by using t=3 hours and A=3000. This means $A_0=6000$ in this context.

$A=A_0e^{rt}\\3000=6000e^{r(3)}0.5=e^{(24.5)r}$

After substituting, we divided by 6000 to each side to get 0.5 on the left. Now to solve for r, we will take the natural log of both sides and use log rules to isolate r.

$ln 0.5=ln e^{(3)r}\\ln 0.5=3r (ln e)\\\frac{ln0.5}{3} =r$

We know$lne=1$ so we were able to cancel it out and divide both sides by 3.

We solve with a calculator

$\frac{ln0.5}{3} =r\\-0.23=r$

We change -0.23 into a percent by multiplying by 100 to get -23% as the rate.

The equation is$A=6000e^{-0.23t}$

Answer 2

Answer:

$f(t) = 6000 (\frac{1}{2})^t^/^3$

Step-by-step explanation:

Let's take the equation:

$y = y_0 e^-^k^t$

Given the initial population of bacteria = 6000,

$y = 6000e^-^k^t$

Population of bacteria after 3 hours is 3000.

y(3) = 3000

Thus,

$6000e^-^k^3 = 3000$

$e^-^k^3 = \frac{3000}{6000}$

$e^-^k^3 = \frac{1}{2}$

$(e^-^k)^3 = \frac{1}{2}$

$e^-^k= (\frac{1}{2})^1^/^3$

Let's substitute $(\frac{1}{2})^1^/^3$ for $e^-^k$ in $y = 6000e^-^k^t$

Therefore, we have:

$6000e^-^k^t$

$= 6000 [(\frac{1}{2}) ^1^/^3]^t$

$= 6000 [\frac{1}{2}] ^t^/^3$