# Exponential Growth

• Aug 16th 2008, 01:18 PM
CalcGeek31
Exponential Growth
Here is the question:
Under ideal conditions, a certain bacteria population is known to double every three hours. Initially there are 100 bacteria.

I need to figure out how many bacteria there are at 15 hours and at t hours... I really need help with this please help.
• Aug 16th 2008, 01:25 PM
galactus
We have 200 bacteria in 3 hours because it doubles.

$\displaystyle 200=100e^{3k}$

Solve for $\displaystyle k=\frac{ln(2)}{3}$

$\displaystyle A=100e^{t\frac{ln(2)}{3}}$

Let t=15.
• Aug 16th 2008, 01:28 PM
CalcGeek31
Okay thank you but now I have to chang it completely to ln/log format to get something = to t right?
• Aug 16th 2008, 02:02 PM
galactus
You can solve for t if your wish, but the equation I gave should suffice for the number of bacteria given any t.
• Aug 16th 2008, 02:08 PM
CalcGeek31
unfortunately I have to solve for t according to my BC Calc Packet
• Aug 16th 2008, 02:14 PM
galactus
You can also express $\displaystyle e^{\frac{ln(2)t}{3}}=2^{\frac{t}{3}}$

$\displaystyle A=100e^{\frac{ln(2)t}{3}}$

$\displaystyle t=\frac{3ln(\frac{A}{100})}{ln(2)}$