Question

A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria and after 8 hours the count is 50,000. a. Find the initial population b. Find an expression for the population after t hours.

Answer 1

Answer:

So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).

Step-by-step explanation:

Given:

A bacteria culture with constant relative rate.

No. of bacteria = $400$        Time ($t$) = $2$ hrs

No.of bacteria = $50,000$    Time ($t$) = 8 hrs

Here the growth is exponential.

Formula to be used.

• $P(t)=P_0e^k^t$ where $P(t)$ = Number of bacteria after time ($t$).

                                        $P_0$ = Initial population

According to the question.

$P(2)=P_0e^2^k=400$                 ...equation (i)

$P(8)=P_0e^8^k =50,000$            ...equation (ii)

Dividing equation (ii) with (i)

⇒ $\frac{P_0e^8^k}{P_0e^2^k} =\frac{50,000}{400}$

⇒ $e^6^k=125$

⇒ $(e^k)^6=125$

Now using $\ln$ function both sides.

⇒ $\ln (e^k)^6=\ln(125)$

⇒ $6\ln(e^k)=\ln (125)$  using $\ln x^a=a\ln x$ rule

⇒ $\ln e^k=\frac{\ln 125}{6}$

And from logarithmic rule $\ln(e) = 1$

⇒ $k=\frac{\n 125}{6}$

⇒ $k=0.8047$

Now plugging this k value in any of the equation we can find the initial population.

a)

Then,

$P_0(e^0^.^8^0^4^7^\times ^2) = 400$

$P_0=80.003$

The initial population of the bacteria culture is 80.003

b)

Expression for the population after $t$ hours.

$P(t)=80.003\times e^0^.^8^0^4^7 ^(^t^)$

So, the initial population of the bacteria culture is 80.003 and the expression after t hours is P(t)=80.003 e^(0.8047\times t).