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Question

(C) Let $x$ be the number of bacteria at time $t$. $\frac{dx}{dt} \propto x \Rightarrow \frac{dx}{dt} = kx \Rightarrow \int \frac{dx}{x} = \int k \cdo

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(C) Let $x$ be the number of bacteria at time $t$. $\frac{dx}{dt} \propto x \Rightarrow \frac{dx}{dt} = kx \Rightarrow \int \frac{dx}{x} = \int k \cdot dt$ $\log x = kt + c$ Initially,i.e.,when $t = 0$,let $x = x_{0}$. $\log x_{0} = k 	imes 0 + c \Rightarrow c = \log x_{0}$ $\log x - \log x_{0} = kt \Rightarrow \log \left(\frac{x}{x_{0}}ight) = kt$ ... $(1)$ Since the number doubles in $4 	ext{ hours}$,i.e.,$t = 4$ and $x = 2x_{0}$: $\log \left(\frac{2x_{0}}{x_{0}}ight) = 4k \Rightarrow \log 2 = 4k \Rightarrow k = \frac{1}{4} \log 2$ When $t = 12$,substituting $k$ into equation $(1)$: $\log \left(\frac{x}{x_{0}}ight) = 12 	imes \left(\frac{1}{4} \log 2ight) = 3 \log 2 = \log 2^{3} = \log 8$ Therefore,$\frac{x}{x_{0}} = 8 \Rightarrow x = 8x_{0}$. Given the original number is $N$,the number after $12 	ext{ hours}$ will be $8N$.

The bacteria increases at a rate proportional to the number of bacteria present. If the original number $N$ doubles in $4 \text{ hours}$,then the number of bacteria in $12 \text{ hours}$ will be (in $N$)

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