Question

gosh I struggle with word problems.

40,000 bacteria per milliliter to start. If start amount triples in 4 days, how many days does it take to reach 72,000?

Answer 1

about 2.4 days

Answer 2

cool, is there an exact amount? I need to be able to round to the appropriate decimal place. :)

Answer 3

oops triples in 4 days !!

Answer 4

\[40,000\times 3^{\frac{x}{4}}=72,000\]solve for \(x\)

Answer 5

you know how to do that?

Answer 6

no. now i'm confused. i thought there was a certain formula involved. n(t)=Noe^0.14(t).
:/ i have no idea how the end result is 2.4 days

Answer 7

@whpalmer4 i have some extra credit problems i'm trying to work on. I'm just so bad at the word problems.

Answer 8

\[40000*3^{x/4}=72000\]Divide both sides by 40000\[\frac{72000}{40000} = 3^{x/4}\]Take log of both sides\[\ln \frac{72}{40} = \frac{x}{4}\ln 3\] Solve for \(x\)\[\frac{4\ln 1.8}{\ln 3} =x \approx 2.14011\]

Answer 9

That formula \[n(t) = N_0 e^{0.14t}\]is equivalent to the other one...just a different base for the exponent.

Answer 10

you can do it the other way too, but it is more work
@whpalmer4 shows how to do it using the easy formula i wrote

Answer 11

i really do struggle with these types of problems and @whpalmer4 was helping me previously. lol he just knows I need a longer explanation. :)

Answer 12

Whoops, I spoke too soon. Where did the 0.14 come from?

Answer 13

i'm not even sure anymore. my brain is so fried from math overload. lol

Answer 14

Okay, here's how you can figure it from the other formula:

We want the population to triple in 4 days. That means \(N(t) = 3N_0\) or\[3N_0 = N_0e^{\lambda t}\]Obviously the initial population doesn't matter, so we just have \[3 = e^{4\lambda}\]\[\ln 3 = 4\lambda\]\[\lambda = \frac{\ln 3}{4} \approx 0.274653\]
So our completed formula with the growth constant (\lambda\) in place is
\[N(t) = N_0e^{\lambda t}\]\[N(t) = 40000 e^{0.274653t}\]

To find the value of \(t\) where \(N(t) = 72000\):
\[72000 = 40000 e^{0.274653t}\]\[1.8 = e^{0.274653t}\]\[\ln 1.8 = 0.274653t\]\[t \approx \frac{0.587787}{0.274653} \approx 2.14011\]

Answer 15

For just being able to eyeball the formula and pick out information of interest, using the doubling or tripling figure as the base makes it easier. And you can just read off (or insert) the time, as it is the denominator of the fraction in the exponent.

Want something that grows by a factor of 5.4 every 3.9 days?
\[N(t) = N_0(5.4)^{t/3.9}\]Bam! You're done!