A colony of germs grows exponentially. It starts at 500 germs. After 8 minutes there are 600 germs. When will there be 1000 germs?
600=500e^a8 where a is a constant.
1000=600e^aX
solve for a in the first equation:
1.2=e^8a
take ln of both sides
8a=.182
a=.02279
now put that in the second equation
1000/600=e^ax
take ln both sides
.511=ax= .02279x
x=22.4 min from when it was 600,or 30.4 min from the start.
check my work
To determine when there will be 1000 germs in the colony, we need to find the time it takes for the colony to grow from 600 germs to 1000 germs.
Let's first calculate the growth rate of the colony. We have the initial number of germs (500) and the number of germs after 8 minutes (600).
We can use the exponential growth formula:
N = N0 * e^(rt)
Where:
N = Final number of germs
N0 = Initial number of germs
r = Growth rate
t = Time in minutes
e = Base of the natural logarithm (approximately 2.71828)
Using the given information, we can plug in the values:
600 = 500 * e^(8r)
Next, let's isolate the exponential term:
e^(8r) = 600/500 = 1.2
Using natural logarithms, we can take the ln of both sides to eliminate the exponential:
ln(e^(8r)) = ln(1.2)
By the logarithmic property, the exponent can be moved to the front:
8r * ln(e) = ln(1.2)
Since ln(e) is equal to 1, we have:
8r = ln(1.2)
Now, we can solve for r:
r = ln(1.2) / 8
By substituting this r value back into the exponential growth formula, we can determine how long it takes for the colony to grow from 600 germs to 1000 germs:
1000 = 600 * e^((ln(1.2)/8) * t)
Divide both sides by 600:
1000/600 = e^((ln(1.2)/8) * t)
Simplifying further:
5/3 = e^((ln(1.2)/8) * t)
To solve for t, we can take the natural logarithm (ln) of both sides:
ln(5/3) = (ln(1.2)/8) * t
Finally, isolate t by dividing both sides by (ln(1.2)/8):
t = (ln(5/3)) / (ln(1.2)/8)
Calculating this expression will give us the time required for the colony to grow from 600 germs to 1000 germs.