Answer:
P = 94800 inh (town population at t = 10)
Step-by-step explanation: We get:
dP/dt = kP (1)
Population in year 2 is 30000
Population in year 4 is 40000
From equation (1)
dP/P = kdt
Then
∫ 1/P dp = k ∫dt
Ln (P) = kt + c (2)
Now we know populations in years 2 and 4 then by subtitution
Ln (40000) = 4k + c
Ln ( 30000) = 2k + c
Subtracting these equation we have
Ln (40000) - Ln ( 30000) = 2k
Then we can get k ??
ln ( 40000/30000) = 2k k = 1/2 Ln ( 4/3)
Now weareable to find population when t = 10
Equation (2)
Ln (P) = kt + c and from previous data
Ln (40000) = 4k + c
ln (P) = 10* k + c
Ln (40000) = 4k + c
The same procedure
Ln (P) - Ln ( 40000) = 6k
Ln (P) - Ln ( 40000) = 6* 1/2 Ln ( 4/3)
Ln (P) - Ln ( 40000) = 3*Ln (4/3)
Ln [ P/ 40000 ] = 3*Ln (4/3)
Eliminating Ln on both sides of the equation
P / 40000 = (4/3)³
P = 40000 * (4/3)³
P = 40000 * 2.37
P = 94800 inh (town population at t = 10)
