Question

The population of fish in a certain lake follows the logistic growth function , where t is the time in years.

Answer 1

Answer:

46 years

Step-by-step explanation:

We have the logistic growth function $f(t)=\frac{25,000}{1+8.25e^{-0.076t}}$  and we want to find the time when the population will reach 20,000, to do it we just need to replace $f(x)$ with 20,000 and solve for $t$:

$f(t)=\frac{25,000}{1+8.25e^{-0.076t}}$

$20,000=\frac{25,000}{1+8.25e^{-0.076t}}$

Divide both sides by 25,000

$\frac{20,000}{25,000} =\frac{1}{1+8.25e^{-0.076t}}$

$0.8=\frac{1}{1+8.25e^{-0.076t}}$

Multiply both sides by $1+8.25e^{-0.076t}$ and divide  them by 0.8

$1+8.25e^{-0.076t}=1.25$

Subtract 1 from both sides

$8.25e^{-0.076t}=0.25$

Divide both sides by 8.25

$e^{-0.076t}=\frac{0.25}{8.25}$

$e^{-0.076t}=\frac{1}{33}$

Take natural logarithm to both sides

$ln(e^{-0.076t})=ln(\frac{1}{33} )$

$-0.076t=ln(\frac{1}{33} )$

Divide both sides by -0.076

$t=\frac{ln(\frac{1}{33} )}{-0.076}$

$t$ ≈ 46

We can conclude that the population will reach 20,000 after 46 years.