Copyright

A species of fish was added to a lake. The population size P(t) of this species can be modeled by...

Question:

A species of fish was added to a lake. The population size {eq}P(t) {/eq} of this species can be modeled by the following exponential function: {eq}P(t)=1500/(1+7e^{-0.4t}) {/eq} where t is the number of years from the time the species was added to the lake.

Find the initial population size of the species and the population size after 9 years. Round your answers to the nearest whole number as necessary.

Exponential function:

Any function of the form {eq}f(x) = ab^x {/eq}, where {eq}a\neq 0,\ \ b>0,\ \ b\neq 1 {/eq}and x is any real number. constant b is called base and the variable x is called exponent.

Answer and Explanation:

The fish population is given by the function P(t) = {eq}P(t)=\frac{1500}{(1+7e^{-0.4t})} {/eq} where t is the number of years from the time the species was added to the lake.

To find initial population, we need to substitute {eq}t = 0. {/eq}

{eq}\begin{align} P(0) &= \frac{1500}{(1+7e^{-0.4(0)})}\\ &=\frac{1500}{8}\\ &=187.5\\ &\approx 186\ \ fishes. \end{align} {/eq}

To find population after 9 years, we need to substitute t =9 in P(t).

{eq}\begin{align} P(9) &= \frac{1500}{(1+7e^{-0.4(9)})} &=1,259.1646 &\approx 1259\ \ fishes. \end{align} {/eq}

Thus the initial population size of the species is 186 and the population size after 9 years is 1259.


Learn more about this topic:

Loading...
Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
105K

Related to this Question

Explore our homework questions and answers library