So, given a simple population continuous growth problem, it seems that the entirety of the internet uses P=P_0e^(rt) where P is the population over time, P_0 is the initial population, r is the percentage of change, and t is time. But this class is asserting that P=P0e^((lna)(t)) where a=1+r, and everything else stays the same. So lets say a population starts at 10,000 with 10 continuous growth rate. What is the population in 10 years? The internet says that P=10,000e^((.10)(10))=27,182 But the class says that P=10,000e^((ln1.10)10)=25,937 (a=1+r which is 1.10) but that equals the usual constant growth of P=P_0a^t=25937 What the heck? What am I missing?

Ashlynn Hale 2022-08-15 Answered
Exponential continuous growth ln a vs. r? Huh?
So, given a simple population continuous growth problem, it seems that the entirety of the internet uses P = P 0 e r t where P is the population over time, P 0 is the initial population, r is the percentage of change, and t is time.
But this class is asserting that P = P 0 e ( ln a ) ( t ) where a = 1 + r, and everything else stays the same.
So lets say a population starts at 10 , 000 with 10 continuous growth rate. What is the population in 10 years?
The internet says that P = 10 , 000 e ( .10 ) ( 10 ) = 27 , 182
But the class says that P = 10 , 000 e ( ln 1.10 ) 10 = 25 , 937 ( a = 1 + r which is 1.10)
but that equals the usual constant growth of P = P 0 a t = 25937
What the heck? What am I missing?
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Answers (1)

Abdiel Murillo
Answered 2022-08-16 Author has 8 answers
Let us start from what everybody agrees on. The population P ( t ) at time t is given by
(1) P ( t ) = P ( 0 ) e λ t ,
where λ is a constant.
Suppose that the population increases by 10 % in one year. Then putting t = 1 in Equation (1), and using P ( 1 ) = 1.1 P ( 0 ), we obtain
1.1 P ( 0 ) = P ( 0 ) e λ .
From this we obtain that 1.1 = e λ . Taking natural logarithms, we find that λ = ln ( 1.1 ). Thus
P ( t ) = P ( 0 ) e ( ln ( 1.1 ) ) ( t ) .
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