Katydid population is modeled by

Question

N(t) = 50 / 1+9e^ -.33t
where t is number ofyears and N(t) is population is thousands at time t.

when will katydid population reach 40,000?

I somehow got a negative log, producing an error so i can't go any farther. Please help.

Answer 1

N = 50/(1+9e^-.33t) call it
N = 50/(1+9e^(-t/3))
50/N = 1+9e^(-t/3)

9e^(-t/3) = 50/N - 1 = (50-N)/N

e^(-t/3) = (50-N)/9N

-t/3 = ln (50-N)/9N

t = -3 ln (50-N)/9N

You are correct. Log of negative number.

Looking at the original equation,

N(0) = 50/(1+9e), about 2.
As t gets large, we approach 50/1 = 50.

So, what if we try

N = 50000/(1+9e^(-t/3))

That will approach N=50000. This means that the stable population is 50000, not just 50.

Now we have

t = -3 ln (50000-N)/9N
t = -3 ln(10000/360000) = -3ln(1/36) = 10.75 years

Answer 2

Actually, we did not have to use that fudge. N is the population in thousands. So, when the population is 40000, N=40.

The original equation works:

t = -3 ln (50-N)/9N
t = -3 ln(50-40)/360 = -3 ln 1/36 = 10.75 years