Question

Help Please!

Answer 1

I need help with the last problem.

Answer 2

The equation will have the form:
\[\large P=P _{0}e^{kt}\]
where k is the growth constant. It is given that after one year:
\[\large \frac{P}{P _{0}}=3\]
therefore the value of k can be found by solving the equation:
\[\large 3=e^{k}\]
Hint: Take natural logs of both sides.

Answer 3

3 = exp(k)
ln (3) = ln exp(k)
k = ln (3)

Answer 4

@kropot72

Answer 5

@zepdrix

Answer 6

@mathmale

Answer 7

I think it should be P = P_0 e^(ln3)t = 3P_0 e^t

Answer 8

Nope. didn't work :/

Answer 9

Do you put the middle one or the last one?

Answer 10

the last one

Answer 11

try the middle one, please

Answer 12

Nope :/

Answer 13

surrender!!

Answer 14

I found it. I got the right answer using this.

Answer 15

My result was:
\[\large P=P _{0}e^{1.0986t}\]

Answer 16

oh!! that 's why!! logistic growth equation.

Answer 17

For the last part, you need to solve the following equation for t:
\[\large 3000=240.e^{1.0986t}\]

Answer 18

3000 = 240*e^(t*ln(3))
3000 = 240*3^t
3^t = 3000/240
3^t = 25/2
t = (2*log 5 - log 2)/(log 3)