Katydid population is modeled by
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.
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
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
To find when the katydid population reaches 40,000, we need to solve the equation N(t) = 40.
The given equation is N(t) = 50 / (1 + 9e^(-0.33t)).
So to solve for t, we can set up the equation as follows:
40 = 50 / (1 + 9e^(-0.33t))
To simplify the equation, we can multiply both sides by (1 + 9e^(-0.33t)):
40(1 + 9e^(-0.33t)) = 50
Now distribute 40 across the terms inside the parentheses:
40 + 360e^(-0.33t) = 50
Subtract 40 from both sides:
360e^(-0.33t) = 10
Divide both sides by 360:
e^(-0.33t) = 10/360
Simplify the right side:
e^(-0.33t) = 1/36
To solve for t, we can take the natural logarithm of both sides:
ln(e^(-0.33t)) = ln(1/36)
Using the property of logarithms, ln(e^x) = x, where x is any real number:
-0.33t = ln(1/36)
To isolate t, divide both sides by -0.33:
t = ln(1/36) / -0.33
Now we can use a calculator to find the value of t. Remember to use the natural logarithm function (ln) on your calculator:
t ≈ 8.03 years
Therefore, the katydid population will reach 40,000 in approximately 8.03 years.