The population of a newly discovered organism can be described by the function p=500(3)^t/10?

Question

The population of a newly discovered organism can be described by the function p=500(3)^t/10 -- where p is the number of organisms and t is the time (in minutes). What do the 3 and 10 signify in the context of this problem? How long does it take the population to double?

If the question stated: The initial concentration of medicine in a patient' system is 50mg/L. Every 2 hours the concentration decreases by 30%. Write the equation to determin the concentration, C, in terms of time, t, in hours.

Would the equation be: C=50(.3)^t/2 or possibly C=50(-.3)^t/2 ? Where you simply plug in a value for t, let's say "6" to find out what the concentration would be at that time?

Answer 1

for an exponential function the equation to use is y=a*b^x
where a is what you start with
b is what is happening to it
and x is how many times it happens

so for p=500(3)^t/10
500 is what you start with
3 is what is happening to it
so 3 means the population is tripling
t/10 is how many times it happens
if t is the time in munutes then it triples in t/10 minutes
if t was 60 then t/10 would be 60/10=6
that means that the population tripled 6 times....in 60 minutes
once every 10 minutes...so the jpopulation must triple evey 10 minutes

how fast does it double
make a table and figure it out by pluging in values for t
after 1 minute-----558
after 2 min------622.9
after 3-----695
after4----775.9
after 5---866
after 6---966.6
after7---1078.8
so it takes 7 minutes to double

Answer 2

Let us start by working out the initial value, the value of p when t=0.

p=500 x (3)^(0/10)
=500 x 3^0
= 500

That means when recordings were first made, there were 500 members of this organism.

This equation is dominated by the 3^t/10 part, which means it is an exponential equation. The 3 and 10 are constraints into how quickly the population grows. For example had there been a number larger than 3, (i.e. 5), then for any arbitrary time h (h>0), the population would be 5^h/10 as opposed to 3^h/10. From the properties of exponential functions, we know what 5^h/10 is larger than 3^h/10 for any h>0.

Because we are dividing by 10, the larger this divisor is, the slower the rate of increase. For example, if the 10 were replaced by 100, it would take t=100 to achieve the population, that would have taken t=10 in the original equation.

To find double the time to get double the population, we work out the population at the beginning (t=0). This, I have shown earlier to be 500, so double that would be 1000.

So into the equation we let p=1000, and solve for t.

1000=500(3)^t/10
2=3^(t/10)

Now take logs. On a standard calculator you have "ln" or "log". It does not matter as long as the same one is used for both sides of the equation. I'll use the natural log, ln.

ln 2=ln [3^(t/10)] (not evaluating ln 2 yet!)

One property of logarithms is: ln (a^b) = b x ln (a)

ln 2 = t /10 x ln 3
ln 2 / ln 3 = t/ 10
t= 10 x ln 2 / ln 3

Put that in calculator/Excel and there's your answer!

Answer 3

The "3" and "10" mean in every "ten" minute, the population will be "tripled".

Since 500 is the initial population, when the population is doubled, we have p =1000
1000 = 500(3)^t/10

Solve for t,
t = 10ln2 / ln3 = 6.3 min

Answer 4

The 3 and the 10 indicate that the population will triple every 10 minutes.

(3)^(t/10) = 2
(t/10)log3 == log2
t = 10log2/log3
t = 6.3093 min.

Answer 5

once you're searching for a ideal call for this genus some alternatives that are evoked are... Serpens Phallicus (Latin for phallic serpent), Serpens Caecus Braccae (Blind trouser snake in Latin), and Draco Irrumabo (Latin for dick dragon).