Problem 115 - Elementary and Intermediate Algebra

Question

Rodent Control. The rodent population in a city is currently estimated at \(30,000 .\) If is expected to double every 5 years, when will the population reach 1 million?

Step-by-Step Solution

Answer

The rodent population will reach 1 million in approximately 26 years.

Step 1: Understanding the Problem

We are given that the initial rodent population is 30,000. It is expected to double every 5 years. We need to determine how many years it will take for this population to reach 1 million.

Step 2: Setting Up the Equation

We start by identifying the growth model. The doubling period is 5 years, so an exponential growth model applies. The population doubles every 5 years, which can be represented as:\[P(t) = P_0 \times 2^{t/5},\]where \(P_0=30000\) is the initial population, \(P(t)\) is the population at year \(t\), and we look for when \(P(t)=1000000\).

Step 3: Substituting Values

Substitute \(P_0 = 30000\) and \(P(t) = 1000000\) into the equation:\[1000000 = 30000 \times 2^{t/5}.\]

Step 4: Solving for t

To solve for \(t\), first divide both sides by 30,000:\[\frac{1000000}{30000} = 2^{t/5}.\]This simplifies to:\[33.33\overline{3} \approx 2^{t/5}.\]

Step 5: Applying Logarithms

Take the logarithm of both sides of the equation to solve for \(t\):\[\log_{10}(33.33\overline{3}) = \log_{10}(2^{t/5}).\]By properties of logarithms, this simplifies to:\[\log_{10}(33.33\overline{3}) = \frac{t}{5} \log_{10}(2).\]

Step 6: Calculating t

Solve for \(t\) by isolating it:\[t = 5 \times \frac{\log_{10}(33.33\overline{3})}{\log_{10}(2)}.\]Use a calculator to find the logarithm values and calculate \(t\):\[t \approx 5 \times \frac{1.52288}{0.30103} \approx 25.56.\]

Step 7: Conclusion

The population will reach 1 million in approximately 26 years (rounding to the nearest whole number).

Key Concepts

exponential functions

Exponential functions describe situations where quantities grow or decay at rates proportional to their current value. In simpler terms, if something grows exponentially, the bigger it gets, the faster it grows. This is why exponential functions are so powerful for modeling population growth, investments, and radioactive decay.

When modeling an exponentially growing population, like the rodents in our original problem, we start with a formula \[ P(t) = P_0 \times 2^{t/T} \]where:

• \( P(t) \) is the population at time \( t \),
• \( P_0 \) is the initial population size,
• \( 2 \) represents the doubling aspect of growth,
• \( T \) is the doubling time, and
• \( t \) is the time elapsed.

Each part of this equation plays a role in determining how fast the quantities grow. By knowing these components, you can predict how populations will change over time.

logarithms

Logarithms are the mathematical tools that allow us to work backwards from exponential equations. They help us solve for exponents when dealing with exponential functions, especially when variables cannot be isolated through simpler algebraic methods. Think of logarithms as the opposite of exponentiation, much like subtraction is the opposite of addition.

For instance, in the rodent problem, we needed to determine how long it would take for the population to reach a million. Through our equation \[ 33.33\overline{3} \approx 2^{t/5} \]applying logarithms helps isolate \( t \). We apply the logarithm to both sides of the equation:\[ \log_{10}(33.33\overline{3}) = \frac{t}{5} \log_{10}(2) \]This tool, the logarithm, allows us to solve residential or real-life problems involving exponential growth such as determining time-spans involved in doubling populations.

doubling time

Doubling time is a concept used frequently alongside exponential growth when a quantity doubles in size over regular intervals. For the rodent problem, the population doubles every 5 years. This means every 5 years, the population size is multiplied by 2, which is a real-life depiction of exponential growth.

To calculate when the population will reach a certain size, understanding how doubling time relates to the exponential function is key. We use this regular interval in our exponential growth model to predict long-term outcomes.

Typically, doubling time \( T \) is plugged directly into the exponential function giving us a cleaner way to express these growth patterns. Knowing the doubling time for any scenario provides critical insight into how quickly changes are occurring, no matter the starting size. Such knowledge is invaluable when preparing for future outcomes based on growth predictions, such as population control or resource management.