#1
Okay so I have a question on a math lab and I have no clue how to even go about starting it.
In an optimal environment, a bacteria population grows exponentially. At noon
in the sewage treatment plant, experimenters introduce specially cultured bacteria
into a barrel full of rich nutrients. We will assume that the bacteria population in
the barrel is modeled by an exponential function.
Let N(t) be the number of bacteria after t days. Then N(t) = Pa^t for some
constants P and a. Measurements indicate that N(2) = 5, 400 and N(6) = 345, 000.
(a) Before working the problem, estimate (guess!) the value of N(4), the number
of bacteria at noon on the fourth day.
(b) Write down two equations for P and a, one when t = 2 and the other when
t = 6.
(c) Use these two equations to compute a and P. Round your values to three
significant digits, but be sure to store the more precise values for further cal-
culations.

So where I'm confused is that I don't know how to get two values from an equation with two variables. And I'm not trying to get the pit to do my homework. I'm not even asking for the answer. I just need a push in the right direction.
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#2
Quote by gratefullydead2
So where I'm confused is that I don't know how to get two values from an equation with two variables. And I'm not trying to get the pit to do my homework. I'm not even asking for the answer. I just need a push in the right direction.

Make one variable a function of the other, then solve for that one. Then you can substitute that value in and solve for the other.
#4
I used to know how to do this. My mind is shot. :/ Sorry.
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'member The Pit of 10'? oH, I 'member!


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#6
N(2) = 5, 400
and
N(t) = Pa^t
so...
N(2) = Pa^2
and
5400 = Pa^2
Following
345, 000 = Pa^6
.
Now you should easily know what to do. Move one of the equations around so that you get P = ???. and just substitute ??? where the P is in the other equation.
#7
you will have to think a little bit about part a, but part b and c are easy (well for me). for b, you just plug in the values of t into N(t). because you have an N value for them both, you plug that in too. so you have:

5400=P(a^2) and 345000=P(a^6)

solve one for P and plug it into the other. In other words:

P=(5400)/(a^2)

Plug that into the other one and solve for a. Then use that value to solve for P by plugging into one of the three equations (i would use the third one since you already have P solved for).

for the rest of part b, just solve for a.

Hope this helps. Any more algebra/calculus problems, feel free to PM me.
"If A is a success in life, then A equals x + y + z. Work = x; y = play; and z = keeping your mouth shut."
--Einstein

"Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe."
--Einstein
#8
Thanks guys. And JKM i might need to haha. I suck at math.
Quote by Smokey Amp
You can take a noob to water, but you can't like make a noob drink. Y'know?



Don't be a fucktard. Only a teenage yeti would think back hair is cool.

Mackin' bitches and taking it back since 1990
#10
Quote by gratefullydead2
Thanks guys. And JKM i might need to haha. I suck at math.

well math is my specialty. i'm in calculus 3 and physics 1 currently, and will be in differential equations and physics 2 next semester. any math problems you have, i'll be glad to help.

or you can refer to that thread above...either route should yield the same results.

good luck.
"If A is a success in life, then A equals x + y + z. Work = x; y = play; and z = keeping your mouth shut."
--Einstein

"Two things are infinite: the universe and human stupidity; and I'm not sure about the the universe."
--Einstein
Last edited by JKMV11 at Oct 25, 2009,