To determine how long it will take for the population of the town to reach 100,000, given a current population of 76,000 and an annual growth rate of 6.2%, we can use the formula for exponential growth:
\[ P(t) = P_0 (1 + r)^t \]
Where:
- \( P(t) \) is the future population,
- \( P_0 \) is the initial population,
- \( r \) is the growth rate,
- \( t \) is the time in years.
We want to find \( t \) when \( P(t) = 100,000 \):
\[ 100,000 = 76,000 (1 + 0.062)^t \]
First, we simplify the equation:
\[ 100,000 / 76,000 = (1 + 0.062)^t \]
\[ \frac{100,000}{76,000} = (1.062)^t \]
\[ \frac{100}{76} = (1.062)^t \]
\[ \frac{25}{19} = (1.062)^t \]
Next, we need to solve for \( t \). To do this, we'll take the natural logarithm of both sides:
\[ \ln\left(\frac{25}{19}\right) = \ln((1.062)^t) \]
Using the property of logarithms that states \( \ln(a^b) = b \ln(a) \):
\[ \ln\left(\frac{25}{19}\right) = t \ln(1.062) \]
Now we solve for \( t \):
\[ t = \frac{\ln\left(\frac{25}{19}\right)}{\ln(1.062)} \]
Let's calculate the values:
\[ \ln\left(\frac{25}{19}\right) \approx \ln(1.3158) \approx 0.2745 \]
\[ \ln(1.062) \approx 0.0602 \]
\[ t = \frac{0.2745}{0.0602} \approx 4.56 \]
So, it will take approximately 4.56 years for the population to reach 100,000.
the population of a town is increasing at a rate of 6.2% per year. The city council believes they will have to add another elementary school when the population reaches 100 000. If there are currently 76 000 people living in the town, how long do they have before the new school will be needed?
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