Answer :
To find when the population of bacteria will exceed 913, we use the growth model given by the equation [tex]\( P(t) = 800 e^{0.04t} \)[/tex].
Here's a step-by-step solution:
1. Understand the Equation:
- [tex]\( P(t) \)[/tex] represents the population at time [tex]\( t \)[/tex].
- The initial population is 800.
- The growth rate is 0.04 per unit of time.
- We are looking to find the time [tex]\( t \)[/tex] when the population exceeds 913.
2. Set Up the Inequality:
- We set up the inequality [tex]\( 800 e^{0.04t} > 913 \)[/tex] because we want the population to exceed 913.
3. Solve for [tex]\( t \)[/tex]:
- First, divide both sides by 800 to isolate the exponential term:
[tex]\[
e^{0.04t} > \frac{913}{800}
\][/tex]
- Calculate the right-hand side:
[tex]\[
\frac{913}{800} = 1.14125
\][/tex]
- To solve for [tex]\( t \)[/tex], we need to take the natural logarithm (ln) of both sides:
[tex]\[
0.04t > \ln(1.14125)
\][/tex]
- Calculate the natural logarithm:
[tex]\[
\ln(1.14125) \approx 0.131
\][/tex]
- Finally, solve for [tex]\( t \)[/tex] by dividing both sides by 0.04:
[tex]\[
t > \frac{0.131}{0.04}
\][/tex]
- Calculate the value of [tex]\( t \)[/tex]:
[tex]\[
t \approx 3.275
\][/tex]
4. Round the Answer:
- We round 3.275 to one decimal place, which gives us [tex]\( t \approx 3.3 \)[/tex].
Therefore, the population will exceed 913 after approximately [tex]\( t = 3.3 \)[/tex] units of time.
Here's a step-by-step solution:
1. Understand the Equation:
- [tex]\( P(t) \)[/tex] represents the population at time [tex]\( t \)[/tex].
- The initial population is 800.
- The growth rate is 0.04 per unit of time.
- We are looking to find the time [tex]\( t \)[/tex] when the population exceeds 913.
2. Set Up the Inequality:
- We set up the inequality [tex]\( 800 e^{0.04t} > 913 \)[/tex] because we want the population to exceed 913.
3. Solve for [tex]\( t \)[/tex]:
- First, divide both sides by 800 to isolate the exponential term:
[tex]\[
e^{0.04t} > \frac{913}{800}
\][/tex]
- Calculate the right-hand side:
[tex]\[
\frac{913}{800} = 1.14125
\][/tex]
- To solve for [tex]\( t \)[/tex], we need to take the natural logarithm (ln) of both sides:
[tex]\[
0.04t > \ln(1.14125)
\][/tex]
- Calculate the natural logarithm:
[tex]\[
\ln(1.14125) \approx 0.131
\][/tex]
- Finally, solve for [tex]\( t \)[/tex] by dividing both sides by 0.04:
[tex]\[
t > \frac{0.131}{0.04}
\][/tex]
- Calculate the value of [tex]\( t \)[/tex]:
[tex]\[
t \approx 3.275
\][/tex]
4. Round the Answer:
- We round 3.275 to one decimal place, which gives us [tex]\( t \approx 3.3 \)[/tex].
Therefore, the population will exceed 913 after approximately [tex]\( t = 3.3 \)[/tex] units of time.
