# Algebra Word Problem

• Aug 15th 2007, 06:00 PM
Ty Durdan
Algebra Word Problem
A colony of bacteria starts with 900 organisms and double every 12 days
It follows that there will be:
how many organisms after 24 days....I assume 1800 right?
how many after 2 days

how should I approach it?
• Aug 15th 2007, 06:10 PM
TKHunny
Quote:

Originally Posted by Ty Durdan
A colony of bacteria starts with 900 organisms and double every 12 days
It follows that there will be:
how many organisms after 24 days....I assume 1800 right?
how many after 2 days

how should I approach it?

Doubles every 12 days. There is a standard exponential version, but sometimes I like just writing it down...

\$\displaystyle 900*(2)^{Days/12}\$

Test it.

\$\displaystyle 900*(2)^{12/12} = 900*2^{1} = 900*2 = 1800\$

Use it.

\$\displaystyle 900*(2)^{24/12} = 900*2^{2} = 900*4 = 3600\$ Oops. You had better check that.

Use it again.

\$\displaystyle 900*(2)^{2/12} = ???\$
• Aug 15th 2007, 06:24 PM
Ty Durdan
So would it be 75 organisms a day... because that would allow for 900 after 12 days, 1800 after 24 days...and 150 after 2 days.
• Aug 16th 2007, 06:01 AM
TKHunny
You're guessing. Use the formula.

You START with 900. Why would there be only 900 still after 12 days?

Start: 900
12 days: 900*2 = 1800
24 days: 2*12 days = 900*2*2 = ??

Ahem...Sorry, I got a little poetic, there. :rolleyes:
• Aug 16th 2007, 06:14 AM
topsquark
Quote:

Originally Posted by TKHunny
\$\displaystyle 900*(2)^{Days/12}\$

You've got a calculator, do the math. If you are working with 2 days, then:
\$\displaystyle 900*(2)^{2/12} \approx 1010.22\$