The size of a bacteria population is given by P=C*e^(kt) Where C is the initial size of the population, k is the growth rate constant and t is time in minutes.
a) If the bacteria in the population double their number every minute, what is the value of k?
I need help please!
P(0)=Ce^(k*0)=C
P(1)=Ce^(k*1)=Ce^k
P(1)/P(0)=2=(ce^k)/C=e^k
Take natural log
ln(2)=ln(e^k)=k
k=ln(2)=0.6931 approx.
Sure! To find the value of k in the equation P = C*e^(kt), we can use the given information that the bacteria in the population double their number every minute.
When a population doubles in size, the final size (P) is twice the initial size (C). So we can write the equation as:
P = 2C
Now, substitute this into the original equation:
2C = C*e^(kt)
Divide both sides of the equation by C:
2 = e^(kt)
To isolate the variable kt, take the natural logarithm (ln) of both sides of the equation:
ln(2) = ln(e^(kt))
The logarithm of a number to a certain base "b" is the exponent to which the base must be raised to obtain the number. Therefore, ln(e^(kt)) simplifies to kt:
ln(2) = kt
Finally, solve for k by dividing both sides of the equation by t:
k = ln(2)/t
In this case, since the time is given in minutes, t would be equal to 1 minute (because the population doubles every minute).
Hence, the value of k is:
k = ln(2)/1 = ln(2).