To find the bacteria population after 18 hours based on exponential decay, we need to follow several steps. We know from the problem that the population declined from 270,000 to 600 over the course of 24 hours.
The formula for exponential decay is:
[tex]\[ N(t) = N_0 \times e^{-kt} \][/tex]
Where:
- [tex]\( N(t) \)[/tex] represents the quantity at time [tex]\( t \)[/tex],
- [tex]\( N_0 \)[/tex] is the initial quantity,
- [tex]\( k \)[/tex] is the decay constant,
- [tex]\( t \)[/tex] is the passage of time.
We can use the given information to find the decay constant [tex]\( k \)[/tex] by plugging in the values we have for 24 hours:
Initial quantity [tex]\( N_0 = 270,000 \)[/tex],
Quantity after 24 hours [tex]\( N(24) = 600 \)[/tex],
Time [tex]\( t = 24 \)[/tex] hours.
Now we rearrange the formula to solve for [tex]\( k \)[/tex]:
[tex]\[ 600 = 270,000 \times e^{-24k} \][/tex]
Divide both sides by 270,000:
[tex]\[ \frac{600}{270,000} = e^{-24k} \][/tex]
[tex]\[ \frac{1}{450} = e^{-24k} \][/tex]
Now solve for [tex]\( k \)[/tex] by taking the natural logarithm of both sides:
[tex]\[ \ln{\left(\frac{1}{450}\right)} = -24k \][/tex]
[tex]\[ k = -\frac{\ln{\left(\frac{1}{450}\right)}}{24} \][/tex]
Next, we find the population at 18 hours using the same formula:
[tex]\[ N(18) = N_0 \times e^{-18k} \][/tex]
Plug in the values:
[tex]\[ N(18) = 270,000 \times e^{-18 \left(-\frac{\ln{\left(\frac{1}{450}\right)}}{24}\right)} \][/tex]
[tex]\[ N(18) = 270,000 \times e^{\frac{18}{24} \ln{450}} \][/tex]
Since [tex]\( e^{\ln{x}} = x \)[/tex], we can simplify the expression:
[tex]\[ N(18) = 270,000 \times 450^{\frac{18}{24}} \][/tex]
[tex]\[ N(18) = 270,000 \times 450^{0.75} \][/tex]
Now, calculate the value of [tex]\( 450^{0.75} \)[/tex] and multiply by 270,000:
[tex]\[ N(18) = 270,000 \times \sqrt[4]{450^3} \][/tex]
Once you compute this value, you would round the result to the nearest whole number to find out how many bacteria would be present after 18 hours.
