
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.

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.

Okay thank you but now I have to chang it completely to ln/log format to get something = to t right?

You can solve for t if your wish, but the equation I gave should suffice for the number of bacteria given any t.

unfortunately I have to solve for t according to my BC Calc Packet

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