To estimate when the population of bacteria will exceed 2006, we start with the given exponential growth model for the population:
[tex]\[ P(t) = 1000 e^{0.12 t} \][/tex]
We need to find the time [tex]\( t \)[/tex] when the population [tex]\( P(t) \)[/tex] exceeds 2006. So, we set up the equation:
[tex]\[ 1000 e^{0.12 t} = 2006 \][/tex]
To solve for [tex]\( t \)[/tex], follow these steps:
1. Divide both sides by 1000 to isolate the exponential term:
[tex]\[
e^{0.12 t} = \frac{2006}{1000}
\][/tex]
2. Simplify the right-hand side:
[tex]\[
e^{0.12 t} = 2.006
\][/tex]
3. Take the natural logarithm of both sides to eliminate the exponential:
[tex]\[
\ln(e^{0.12 t}) = \ln(2.006)
\][/tex]
4. Use the property of logarithms [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[
0.12 t = \ln(2.006)
\][/tex]
5. Solve for [tex]\( t \)[/tex] by dividing both sides by 0.12:
[tex]\[
t = \frac{\ln(2.006)}{0.12}
\][/tex]
Given the previously determined value of [tex]\(\ln(2.006) / 0.12\)[/tex], we find:
[tex]\[ t \approx 5.8012 \][/tex]
Therefore, the population of bacteria will exceed 2006 at approximately [tex]\( t = 5.8012 \)[/tex] time units.
