Respuesta :
first is to solve the value of k
when A0 = 10000
a(t) = 2000
t = 5
using the formula
a(t) = A0 e^(kt)
2000 = 10000 e^(5k)
2000 / 10000 = e^(5k)
ln ( 1/5) = 5k
k = (ln(1/5)) / 5
k = -1.6094 /5
k = -0.3219
now solve for t,
500 = 10000 e^(-0.3219t)
ln( 500/10000) = -0.3219t
t = 9.3067 years
since that is calculated as the basis at 5 years ago
then actual t is 4.3067 years
when A0 = 10000
a(t) = 2000
t = 5
using the formula
a(t) = A0 e^(kt)
2000 = 10000 e^(5k)
2000 / 10000 = e^(5k)
ln ( 1/5) = 5k
k = (ln(1/5)) / 5
k = -1.6094 /5
k = -0.3219
now solve for t,
500 = 10000 e^(-0.3219t)
ln( 500/10000) = -0.3219t
t = 9.3067 years
since that is calculated as the basis at 5 years ago
then actual t is 4.3067 years
Answer: 4.3 years
Step-by-step explanation:
Given: The population of a species of starfish in the Gulf of Mexico is decreasing at an exponential rate
To find the population we have given an equation [tex]A(t) = A_0e{kt}[/tex], where k is the decay rate (k<0) in t years and[tex]A_0[/tex] is the initial population.
Five years ago the population was 10,000, in 2015 it is only 2000.
So put t=5 , [tex]A_0=10,000[/tex] and A(t)= 2000 in the above function we get
[tex]2000=10,000e^{5k}\\\\\Rightarrow e^{5k}=0.2[/tex]
Taking log on both sides, we get
[tex]5k=\log(0.2)\\\\\Rightarrow5k=-0.698970004336\\\\\Rightarrow\ k=-0.139794000867[/tex]
Taking 2015 as the initial year [tex]A_0=2,000[/tex], To find t when A(t)=500
[tex]500=2000e^{-0.139794000867t}\\\\\Rightarrow e^{-0.139794000867t}=0.25[/tex]
Taking log on both sides, we get
[tex]-0.139794000867t=\log(0.25)\\\\\Rightarrow-0.139794000867t=-0.602059991328\\\\\Rightarrow t=\frac{-0.602059991328}{-0.139794000867}\\\\\Rightarrow t=4.30676558074\approx4.3\ years[/tex]
Hence, This event will occur in 4.3 years from 2015.