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(B) Let $b$ be the number of bacteria. We have $\frac{db}{dt} \propto b \Rightarrow \int \frac{db}{b} = \int K dt$. $	herefore \log b = Kt + c$ ...$(

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$4$ times the original

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(B) Let $b$ be the number of bacteria. We have $\frac{db}{dt} \propto b \Rightarrow \int \frac{db}{b} = \int K dt$. $	herefore \log b = Kt + c$ ...$(1)$ Let $b_{0}$ be the initial number of bacteria. At $t = 0, b = b_{0}$. $\log b_{0} = K(0) + c \Rightarrow c = \log b_{0}$. $	herefore \log \left(\frac{b}{b_{0}}ight) = Kt$ ...$(2)$ When $t = 3, b = 2b_{0}$. $	herefore \log \left(\frac{2b_{0}}{b_{0}}ight) = 3K \Rightarrow K = \frac{1}{3}(\log 2)$. Thus,$\log \left(\frac{b}{b_{0}}ight) = \frac{1}{3}(\log 2)t$. When $t = 6$: $\log \left(\frac{b}{b_{0}}ight) = \frac{1}{3}(\log 2)(6) = 2 \log 2 = \log 4$. $	herefore \frac{b}{b_{0}} = 4 \Rightarrow b = 4b_{0}$. Therefore,the number of bacteria increases $4$ times the original amount.

The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number doubles in $3$ hours,then the number of times the bacteria are increased in $6$ hours is

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