To determine when the population of bacteria will exceed 1114, we use the given growth equation [tex]\( P(t) = 1000 e^{0.03 t} \)[/tex].
Here's the detailed step-by-step solution:
1. Set up the inequality:
We need to find the time [tex]\( t \)[/tex] when the population [tex]\( P(t) \)[/tex] is greater than 1114. This can be expressed as:
[tex]\[
1000 e^{0.03 t} > 1114
\][/tex]
2. Isolate the exponential term:
Divide both sides of the inequality by 1000 to isolate the exponential term:
[tex]\[
e^{0.03 t} > \frac{1114}{1000}
\][/tex]
3. Simplify the fraction:
Simplify the fraction on the right side:
[tex]\[
e^{0.03 t} > 1.114
\][/tex]
4. Apply the natural logarithm:
To solve for [tex]\( t \)[/tex], take the natural logarithm (ln) of both sides:
[tex]\[
\ln(e^{0.03 t}) > \ln(1.114)
\][/tex]
5. Simplify using logarithm properties:
The natural logarithm and the exponential function are inverse functions, so:
[tex]\[
0.03 t > \ln(1.114)
\][/tex]
6. Divide by the coefficient of [tex]\( t \)[/tex]:
Finally, divide both sides by 0.03 to solve for [tex]\( t \)[/tex]:
[tex]\[
t > \frac{\ln(1.114)}{0.03}
\][/tex]
Now, calculating the value on the right side:
[tex]\[
t > \frac{0.1072}{0.03} \approx 3.5986
\][/tex]
Therefore, the population will exceed 1114 when [tex]\( t \approx 3.5986 \)[/tex].
Thus,
[tex]\[ t \approx 3.5986 \][/tex]
The population will exceed 1114 at approximately [tex]\( t = 3.5986 \)[/tex] time units.
