To determine how many years it will take for the population of Honduras to grow from 7,600,000 to 10,000,000 people at an annual growth rate of 2%, compounded continuously, we can use the formula for continuous compound growth:
[tex]\[ P(t) = P_0 \cdot e^{rt} \][/tex]
Where:
- [tex]\( P(t) \)[/tex] is the final population,
- [tex]\( P_0 \)[/tex] is the initial population,
- [tex]\( r \)[/tex] is the growth rate,
- [tex]\( t \)[/tex] is the time in years,
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
Given:
- [tex]\( P_0 = 7,600,000 \)[/tex]
- [tex]\( P(t) = 10,000,000 \)[/tex]
- [tex]\( r = 0.02 \)[/tex] (as a decimal for 2%)
We need to find the number of years [tex]\( t \)[/tex] until the population reaches 10,000,000. First, we rearrange the formula to solve for [tex]\( t \)[/tex]:
[tex]\[ 10,000,000 = 7,600,000 \cdot e^{0.02t} \][/tex]
To isolate [tex]\( t \)[/tex], we divide both sides by 7,600,000:
[tex]\[ \frac{10,000,000}{7,600,000} = e^{0.02t} \][/tex]
Simplifying the fraction on the left side:
[tex]\[ \frac{10,000,000}{7,600,000} = \frac{10}{7.6} \approx 1.3158 \][/tex]
So we have:
[tex]\[ 1.3158 = e^{0.02t} \][/tex]
Next, we take the natural logarithm (ln) of both sides to solve for [tex]\( t \)[/tex]:
[tex]\[ \ln(1.3158) = \ln(e^{0.02t}) \][/tex]
Since [tex]\( \ln(e^x) = x \)[/tex], this simplifies to:
[tex]\[ \ln(1.3158) = 0.02t \][/tex]
To solve for [tex]\( t \)[/tex], we divide both sides by 0.02:
[tex]\[ t = \frac{\ln(1.3158)}{0.02} \][/tex]
Using a calculator to compute the natural logarithm and the division:
[tex]\[ t \approx \frac{0.2744}{0.02} = 13.72 \][/tex]
Therefore, the number of years until the population of Honduras reaches 10,000,000 people is approximately 13.72 years.
So, the answer is:
Blank 1: 13.72
