1. **Problem statement:** We want to find the number of years it takes for a country's population to double given an annual growth rate of $p\%$. 2. **Formula used:** The population growth can be modeled by the exponential growth formula: $$ P(t) = P_0 (1 + \frac{p}{100})^t $$ where $P_0$ is the initial population, $t$ is the number of years, and $p$ is the growth rate percentage. 3. **Doubling condition:** We want $P(t) = 2P_0$, so: $$ 2P_0 = P_0 (1 + \frac{p}{100})^t $$ Dividing both sides by $P_0$: $$ 2 = (1 + \frac{p}{100})^t $$ 4. **Solving for $t$:** Take the natural logarithm of both sides: $$ \ln 2 = t \ln \left(1 + \frac{p}{100}\right) $$ So, $$ t = \frac{\ln 2}{\ln \left(1 + \frac{p}{100}\right)} $$ 5. **Calculate for $p=2$:** Given $\ln 2 \approx 0.6931$ and $\ln 1.02 \approx 0.0198$, $$ t = \frac{0.6931}{0.0198} \approx 35 $$ years (rounded to whole number). 6. **Calculate for $p=3$:** Given $\ln 1.03 \approx 0.02956$, $$ t = \frac{0.6931}{0.02956} \approx 23 $$ years (rounded to whole number). **Final answers:** - For $p=2\%$, population doubles in approximately 35 years. - For $p=3\%$, population doubles in approximately 23 years.