Answer :
After 1 hour, there are [tex]\(729000} \)[/tex] bacteria.
To determine how many bacteria there are after 1 hour, given the initial and growth conditions provided, we will use the exponential growth formula:
[tex]\[ N(t) = N_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( N(t) \)[/tex] is the number of bacteria at time [tex]\( t \)[/tex]
- [tex]\( N_0 \)[/tex] is the initial number of bacteria,
- [tex]\( k \)[/tex] is the growth rate constant.
Given:
- Initial number of bacteria,[tex]\( N_0 = 1000 \).[/tex]
- Number of bacteria after 10 minutes (which is [tex]\( \frac{1}{6} \)[/tex] hours), [tex]( N\left(\frac{1}{6}\right) = 3000 \).[/tex]
Step 1: Find the growth rate constant [tex]\( k \).[/tex]
Using the given data at [tex]\( t = \frac{1}{6} \)[/tex] hours:
[tex]\[ N\left(\frac{1}{6}\right) = 1000 \cdot e^{k \cdot \frac{1}{6}} = 3000 \][/tex]
Divide both sides by 1000:
[tex]\[ e^{k \cdot \frac{1}{6}} = 3 \][/tex]
Take the natural logarithm of both sides to solve for [tex]\( k \):[/tex]
[tex]\[ k \cdot \frac{1}{6} = \ln(3) \][/tex]
Multiply both sides by 6:
[tex]\[ k = 6 \cdot \ln(3) \][/tex]
Step 2: Calculate the number of bacteria after 1 hour (which is 1 hour = 6 times 10 minutes).
Now use [tex]\( t = 1 \)[/tex] hour in the exponential growth formula:
[tex]\[ N(1) = 1000 \cdot e^{k \cdot 1} \][/tex]
[tex]\[ N(1) = 1000 \cdot e^{6 \cdot \ln(3)} \][/tex]
[tex]\[ N(1) = 1000 \cdot (e^{\ln(3)})^6 \][/tex]
[tex]\[ N(1) = 1000 \cdot 3^6 \][/tex]
Calculate[tex]\( 3^6 \):[/tex]
[tex]\[ 3^6 = 729 \][/tex]
Therefore,
[tex]\[ N(1) = 1000 \cdot 729 = 729000 \][/tex]
