Mathematics, 11.03.2020 04:48 Donlito6997

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.

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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 () = hrs

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

Here the growth is exponential.

Formula to be used.

where

= Number of bacteria after time (

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.

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

Expression for the population after 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).

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Answer from: Quest

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A bacteria culture grows with a constant relative growth rate. After 2 hours there are 400 bacteria...

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