# need help growth problem

• June 12th 2009, 10:20 AM
justin016
need help growth problem
given y=Ce^kt where C is the initial value

The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

What is the initial size of the culture?

• June 12th 2009, 11:03 AM
apcalculus
Quote:

Originally Posted by justin016
given y=Ce^kt where C is the initial value

The count in a bacteria culture was 5000 after 15 min and 40000 after 1 hour

What is the initial size of the culture?

Use the two conditions to solve for C and k

Is time measured in hours or minutes?

5000 = C e^(k15)
40000 = C e^(k60)

Divide the second by the first, side by side:
8 = e^(60k-15k)

8 = e^(45k)
ln 8 = 45k

k =ln8 /45

Plug into one of the conditions above to find C. C is the original population, because it is the y-value at t=0.
Good luck!!
• June 12th 2009, 11:56 AM
justin016
i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

The answer i am looking for is 2500. What did i do wrong?
• June 12th 2009, 01:10 PM
apcalculus
Quote:

Originally Posted by justin016
i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

The answer i am looking for is 2500. What did i do wrong?

Try units in hours. I assumed minutes in the notes above.
• June 12th 2009, 02:58 PM
justin016
Quote:

Originally Posted by apcalculus
Try units in hours. I assumed minutes in the notes above.

I still get the same answer
• June 12th 2009, 08:50 PM
mr fantastic
Quote:

Originally Posted by justin016
i plugged k= ln8/45 into Ce^k15=5000 to find C ,and i get 12.2877

The answer i am looking for is 2500. What did i do wrong?

If you don't post your working, how can you expect anyone to answer this? And how can you expect to learn from your mistake(s).

Quote:

Originally Posted by justin016
I still get the same answer

A start:

$5000 = C e^{\left( \frac{\ln 8}{45}\right) 15} = C e^{\frac{\ln 8}{3}} = C e^{\ln 2}$

where you get that last expression by applying a well known log rule.
• June 13th 2009, 04:44 PM
justin016
Ce^ln8/3=5000
Ce^ln2=5000
C2=5000
C=5000/2
C=2500

I got confuse where you changed from ln8/3=ln2. How did you do that?
• June 13th 2009, 05:23 PM
mr fantastic
Quote:

Originally Posted by justin016
Ce^ln8/3=5000
Ce^ln2=5000
C2=5000
C=5000/2
C=2500

I got confuse where you changed from ln8/3=ln2. How did you do that?

$\frac{\ln 8}{3} = \frac{1}{3} \ln 8 = \ln 8^{1/3}$ using the usual log rule $a \ln b = \ln b^a$.

And you should know that $8^{1/3} = \sqrt[3] 8 = 2$.
• June 13th 2009, 05:26 PM
justin016
Quote:

Originally Posted by mr fantastic
$\frac{\ln 8}{3} = \frac{1}{3} \ln 8 = \ln 8^{1/3}$ using the usual log rule $a \ln b = \ln b^a$.

And you should know that $8^{1/3} = \sqrt[3] 8 = 2$.

Okay i got it, thanx alot