Question

If the population grows at the rate of $8\%$ per year, then the time taken for the population to be doubled is (Given $\log 2 = 0.6912$)

• A. $6.8\ \text{years}$
• B. $4.3\ \text{years}$
• C. $10.27\ \text{years}$
• D. $8.64\ \text{years}$

Hint:

For any continuous growth problem modeled by $\frac{dP}{dt} = rP$, the doubling time is always given by the direct formula $t = \frac{\log 2}{r}$. Simply dividing the given logarithm value by the decimal rate ($0.08$) skips the integration step entirely!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question describes a population growth process modeled by a first-order differential rate. The growth rate is constant at $8\%$ per year, and we need to compute the total time required for the initial population to double.

Step 2: Key Formula or Approach:
Let $P$ represent the population at any time $t$. The rate of change of population with respect to time is proportional to the population present: $$\frac{dP}{dt} = 8\% \text{ of } P = \frac{8}{100}P = 0.08P$$ We solve this differential equation by separating variables and integrating: $$\int \frac{1}{P} \, dP = \int 0.08 \, dt$$

Step 3: Detailed Explanation:
Integrating both sides of the equation yields: $$\log P = 0.08t + C$$ Let the initial population at $t = 0$ be $P_0$. Substituting these values allows us to solve for the integration constant $C$: $$\log P_0 = 0.08(0) + C \implies C = \log P_0$$ Substitute $C$ back into our integrated equation: $$\log P = 0.08t + \log P_0 \implies \log P - \log P_0 = 0.08t \implies \log\left(\frac{P}{P_0}\right) = 0.08t$$ We want to find the time $t$ when the population doubles, which means $P = 2P_0$: $$\log\left(\frac{2P_0}{P_0}\right) = 0.08t \implies \log 2 = 0.08t$$ Isolate time $t$ using the provided value for $\log 2 = 0.6912$: $$t = \frac{\log 2}{0.08} = \frac{0.6912}{0.08} = \frac{69.12}{8} = 8.64\ \text{years}$$

Step 4: Final Answer:
The time taken for the population to double is $8.64\ \text{years}$, corresponding to option (D).