Question

How many bacteria will be present after 5 hours if a culture of bacteria obeys the law of uninhibited growth where if 500 bacteria are present in the culture initially and there are 800 after 1 hour?

Answer 1

Hint: Here we will use the basic of the differential equation for the law of uninhibited growth. Then we will integrate the equation to get it free from differential function. Then we will use the given initial conditions to find the value of the constant. Then by using this equation we will get the value of the number of bacteria that will be present after 5 hours.

Complete step by step solution:
Given that the culture of bacteria obeys the law of uninhibited growth.
We know that the law of uninhibited growth is according to the differential equation. Therefore, we get
\[\dfrac{{dN}}{{dt}} = kN\]
Now we will integrate the equation to remove the differential from the equation. Therefore after integration we get
 \[ \Rightarrow \ln \left( {\dfrac{{{N_2}}}{{{N_1}}}} \right) = k\left( {{t_2} - {t_1}} \right)\]……………….. \[\left( 1 \right)\]
Now we will use the initial given condition to get the value of the constant \[k\]. Therefore the initial conditions are
At \[{t_1} = 0\] the number of bacteria is \[{N_1} = 500\]
At \[{t_2} = 1\] the number of bacteria is \[{N_2} = 800\]
So by putting these values in the equation \[\left( 1 \right)\], we get
\[ \Rightarrow \ln \left( {\dfrac{{800}}{{500}}} \right) = k\left( {1 - 0} \right)\]
\[ \Rightarrow k = \ln \left( {\dfrac{8}{5}} \right)\]
Therefore putting the value of \[k\] in the equation \[\left( 1 \right)\], we get
\[ \Rightarrow \ln \left( {\dfrac{{{N_2}}}{{{N_1}}}} \right) = \ln \left( {\dfrac{8}{5}} \right)\left( {{t_2} - {t_1}} \right)\]
Now we have to find the number of bacteria will be present after 5 hours. Therefore using the initial conditions are
At \[{t_1} = 0\] the number of bacteria is \[{N_1} = 500\]
At \[{t_3} = 5\] the number of bacteria is \[{N_3}\]
So putting these values in the above equation to get the value of \[{N_3}\]. Therefore, we get
\[ \Rightarrow \ln \left( {\dfrac{{{N_3}}}{{500}}} \right) = \ln \left( {\dfrac{8}{5}} \right)\left( {5 - 0} \right)\]
\[ \Rightarrow \ln \left( {\dfrac{{{N_3}}}{{500}}} \right) = 5\ln \left( {\dfrac{8}{5}} \right) = 2.35\]
Now we will apply the exponential to both side of the equation, we get
\[ \Rightarrow {e^{\ln \left( {\dfrac{{{N_3}}}{{500}}} \right)}} = {e^{2.35}}\]
The exponential and the logarithmic function will cancel out. Therefore, we get
\[ \Rightarrow \dfrac{{{N_3}}}{{500}} = 10.485\]

\[ \Rightarrow {N_3} = 10.485 \times 500 \simeq 5243\]
Hence the number of bacteria will be present after 5 hours if a culture of bacteria obeys the law of uninhibited growth are 5243.

Note:
Here we should know the basic differential equation for Newton's law of growth which is the equation for the rate of growth of the substance. Whenever the term rate is attached to anything it varies with the time which means it varies i.e. increases or decreases with respect to the time. When the rate of growth is decreasing then there is a negative sign in the equation and there is a positive sign when the rate of growth is increasing. We should very carefully apply the initial conditions to the equation of the rate of growth to get the value of the constant of the equation.