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(B) Let $x$ be the number of bacteria in a certain culture at time $t$. The rate of increase is $\frac{dx}{dt}$,which is proportional to $x$. $\frac{d

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(B) Let $x$ be the number of bacteria in a certain culture at time $t$. The rate of increase is $\frac{dx}{dt}$,which is proportional to $x$. $\frac{dx}{dt} = Kx \Rightarrow \frac{dx}{x} = Kdt$. Integrating both sides,we get $\ln x = Kt + C$. At $t = 0$,let $x = x_0$,so $C = \ln x_0$. Thus,$\ln \left(\frac{x}{x_0}\right) = Kt$. Given that the number doubles in $4$ hours,at $t = 4$,$x = 2x_0$. $\ln(2) = 4K \Rightarrow K = \frac{\ln 2}{4}$. Substituting $K$ back,$\ln \left(\frac{x}{x_0}\right) = \frac{t}{4} \ln 2$. For $t = 12$,$\ln \left(\frac{x}{x_0}\right) = \frac{12}{4} \ln 2 = 3 \ln 2 = \ln(2^3) = \ln 8$. Therefore,$\frac{x}{x_0} = 8$,which means the bacteria increase by $8$ times.

In a certain culture of bacteria,the rate of increase is proportional to the number present. It is found that the number doubles in $4$ hours. Then the number of times the bacteria are increased in $12$ hours is

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