# Illegal Fish

Alignments to Content Standards: F-LE.B.5 F-LE.A.1.c

A ﬁsherman illegally introduces some ﬁsh into a lake, and they quickly propagate. The growth of the population of this new species (within a period of a few years) is modeled by $P(x)=5b^x$, where $x$ is the time in weeks following the introduction and $b$ is a positive unknown base.

1. Exactly how many fish did the fisherman release into the lake?

2. Find $b$ if you know the lake contains 33 fish after eight weeks. Show step-by-step work.

3. Instead, now suppose that $P(x)=5b^x$ and $b=2$. What is the weekly percent growth rate in this case? What does this mean in every-day language?

## IM Commentary

This is a direct task suitable for the early stages of learning about exponential functions. Students interpret the relevant parameters in terms of the real-world context and describe exponential growth.

## Solution

1. The fisherman released the fish into the lake at time zero, $t=0$, the exact moment of introduction. Thus, the number of fish that the fisherman released into the lake is given by:

\begin{align} P(0) &= 5b^0 \\ P(0) &= 5 \cdot 1 \\ P(0) &= 5 \end{align}

This means that the fisherman released 5 fish into the lake.

2. We know that $x$ is the time in weeks following the introduction. Let us assume that 2 months is approximately 8 weeks, giving $t=8$. Then, if the lake contains 33 fish after two months, or $P(8)=33$, we can solve for $b$:

\begin{align} 33 &= 5b^8 \\ b^8 &= \frac{33}{5} \\ b &= \left( \frac{33}{5} \right)^{\frac18} \\ b &\approx 1.266 \end{align}

Thus, $b$ is approximately equal to 1.2 if the lake contains 33 fish after two months.

3. The “weekly percent growth rate” is the percent increase of the population in one week. Since $b=2$, we know that the population at any time $x$ is given by $P(x)=5\cdot 2^x$, and that the population one week later is given by $$P(x+1)=5\cdot 2^{x+1}=(5\cdot 2^x)\cdot 2=2P(x).$$ We learn that the population doubles each week, which is to say that there is a 100% weekly growth rate.