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).