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(D) Let the initial population be $P$ and the rate of growth be $r = 10 \% = 0.1$. Assuming continuous growth,the population $P(t)$ at time $t$ is giv

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None of the Above

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(D) Let the initial population be $P$ and the rate of growth be $r = 10 \% = 0.1$. Assuming continuous growth,the population $P(t)$ at time $t$ is given by $P(t) = P_0 e^{rt}$. For the population to double,$P(t) = 2P_0$. $2P_0 = P_0 e^{0.1t} \Rightarrow 2 = e^{0.1t}$. Taking the natural logarithm on both sides: $\ln 2 = 0.1t$. $t = \frac{\ln 2}{0.1} = 10 \ln 2 	ext{ years}$. If the growth is compounded annually,$2P = P(1 + 0.1)^n \Rightarrow 2 = (1.1)^n$. Taking $\log_{10}$ on both sides: $\log 2 = n \log 1.1$. $n = \frac{\log 2}{\log 1.1} \approx \frac{0.3010}{0.0414} \approx 7.27 	ext{ years}$. Since the result $10 \ln 2$ or $7.27$ does not match the options provided,the correct choice is $D$.

$A$ population grows at the rate of $10 \%$ of the population per year. How long does it take for the population to double?

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