1. ## Exponential equation help

Can someone help tell me how to do this? Thanks in advance.

At any time t>0, the growth of a bacteria culture is growing exponentially and y represents the amount of bacteria present at any time t. The initial population (t=0) is 1000 and the population triples during the first 5 days. Write an expression for y at any time t≥0.

2. Sorry for the double post, but could someone help me?

I got an answer of (4.5735^x)+999. Is that correct?

3. Originally Posted by Cursed
Can someone help tell me how to do this? Thanks in advance.

At any time t>0, the growth of a bacteria culture is growing exponentially and y represents the amount of bacteria present at any time t. The initial population (t=0) is 1000 and the population triples during the first 5 days. Write an expression for y at any time t≥0.
$\displaystyle y = y_0e^{\lambda t}$

We know that $\displaystyle y_0 = 1000$ and that in t = 5 days the population triples. So
$\displaystyle 3000 = 1000e^{5 \lambda}$

$\displaystyle 3 = e^{5\lambda}$

$\displaystyle ln(3) = 5 \lambda$

$\displaystyle \lambda = \frac{ln(3)}{5} \approx 0.219722$

So
$\displaystyle y = 1000e^{0.219722 t}$

This is an acceptable answer, but we can do a bit better if you wish to do a little more algebra.

Recall $\displaystyle \lambda = \frac{ln(3)}{5}$

So
$\displaystyle y = 1000e^{t \cdot ln(3)/5}$

Using exponent properties:
$\displaystyle y = 1000 \left ( e^{ln(3)} \right ) ^{t/5}$

$\displaystyle y = 1000 \cdot 3^{t/5}$

-Dan