Exponential question - please help?

Question

x=Ae^-kt

where A and k are constants

If x=0.2 when t=0 and x=0.05 when t=3 what are A and k???
(please explain how to work this out as well as the answer)


Find x when t=4???

Answer 1

do your own homework

Answer 2

Sorry! but i forgot the exact value of e, it something 3.xx

ok here goes.
if t= 0 then
= x=Ae^-kt => 0.2=Ae^0 thus the it will be 0.2=Ae => 0.2/e =A

now u got the value of constant A
put it in the next problem and find the value of k

Answer 3

0.2 = Ae^0 = A
So A=0.2
0.05 = 0.2 e^(-3k)
==> e^3k = 4
3k = ln 4
k = (ln 4)/3 = 0.46209812
At t=4
x = 0.2 e^( - 0.46209812 *4) = 0.031498026

Answer 4

x=Ae^-kt ......(1)
when x=0.2,t=0
substitute into (1)
0.2=A*1
>>>A=0.2

x=Ae^-kt
take ln of each side;
lnx=lnA+ln(e^-kt)
lnx=lnA-kt
k=(lnA-lnx)/t
when x=0.05,t=3
k=((ln(0.2)-ln(0.05))/3
=0.4620981204
substitute into (1)
x=0.2*e^(-0.4620981204*t)
{A=0.2,k=0.4620981204}

when t=4,
x=0.2*e^(-0.4620981204*4)
=0.0314980262

i hope that this helps

Answer 5

0.2 = A
0.05 = Ae^(- 3k)
0.05 = 0.2 e^(- 3k)
5 = 20 e^(- 3k)
0.25 = e^(- 3k)
log 0.25 = - 3k
k = log 0.25 / (-3) = 0.46
x = (0.2).e^(- 0.46 x 4)
x = (0.2).e^(- 1.84)
x = 0.481