# Exponential equation help

• Oct 11th 2007, 02:22 PM
Cursed
Exponential equation help
Can someone help tell me how to do this? Thanks in advance. :D

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.
• Oct 11th 2007, 03:37 PM
Cursed
Sorry for the double post, but could someone help me?

I got an answer of (4.5735^x)+999. Is that correct?
• Oct 11th 2007, 06:37 PM
topsquark
Quote:

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

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.

$y = y_0e^{\lambda t}$

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

$3 = e^{5\lambda}$

$ln(3) = 5 \lambda$

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

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

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

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

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

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

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

-Dan