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Rate of growth of bacteria is proportional to the number of bacteria present at that time. If $x$ is the number of bacteria present at any instant $t$, then which one of the following is correct? (Take proportional constant equal to 1)
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Verified Answer
The correct answer is:
$x=c e^{t}$
Rate of growth of bacteria $\propto$ number of bacteria present at that time
$\Rightarrow \quad \frac{d x}{d t} \propto x \Rightarrow \frac{d x}{d t}=x(\because$ Proportional constant $=1)$
$$
\begin{array}{l}
\Rightarrow \quad \int \frac{1}{x} d x=\int d t \\
\Rightarrow \quad \log x=t+\log c \\
\Rightarrow \quad \log x-\log c=t \\
\Rightarrow \quad \log \left(\frac{x}{c}\right)=t \\
\Rightarrow \quad \frac{x}{c}=e^{t} \\
x=c e^{t}
\end{array}
$$
$\Rightarrow \quad \frac{d x}{d t} \propto x \Rightarrow \frac{d x}{d t}=x(\because$ Proportional constant $=1)$
$$
\begin{array}{l}
\Rightarrow \quad \int \frac{1}{x} d x=\int d t \\
\Rightarrow \quad \log x=t+\log c \\
\Rightarrow \quad \log x-\log c=t \\
\Rightarrow \quad \log \left(\frac{x}{c}\right)=t \\
\Rightarrow \quad \frac{x}{c}=e^{t} \\
x=c e^{t}
\end{array}
$$
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