# Algebra Word Problem

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• August 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?
• August 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...

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

Test it.

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

Use it.

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

Use it again.

$900*(2)^{2/12} = ???$
• August 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.
• August 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?

Think about doubling.

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

Seriously, WRITE DOWN a working formulation. Don't try to do it in your head or on your calculator. Let the notation help you. Allow the algebra to soak into your brain and pour over your soul...:eek:

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

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

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

So about 1010 organisms.

-Dan