Exponential and Logarithmic Functions: Applications in Business Calculus

Exponential and Logarithmic Functions

Introduction

Exponential and logarithmic functions are fundamental in Business Calculus, providing mathematical models for growth and decay processes in various fields such as finance, physical sciences, and life sciences. This section focuses on the application of exponential decay, including the concepts of decay rate, half-life, and their use in real-world scenarios.

Exponential Decay

Definition and Differential Equation

Exponential decay describes a process where a quantity decreases at a rate proportional to its current value. The general form of the differential equation for exponential decay is:

• Differential Equation: $\frac{dP}{dt} = -kP$, where $k > 0$

• Solution: $P(t) = P_0 e^{-kt}$

• Interpretation: $P_0$ is the initial amount at time $t = 0$; $k$ is the decay rate.

This equation models the exponential decrease of $P$ over time.

Exponential Growth vs. Decay

Exponential functions can model both growth and decay, depending on the sign of the exponent:

• Growth: $P = P_0 e^{kt}$ for $k > 0$

• Decay: $P = P_0 e^{-kt}$ for $k > 0$

Graphically, growth curves rise rapidly, while decay curves fall quickly from the initial value.

Half-Life and Decay Rate

Relationship Between Decay Rate and Half-Life

The half-life ($T$) of a substance is the time required for it to decrease to half its initial amount. The decay rate ($k$) and half-life are related by:

• $T = \frac{\ln 2}{k}$

• $k = \frac{\ln 2}{T}$

This relationship is essential for converting between the rate of decay and the time it takes for a substance to halve.

Example: Physical Science – Half-life

For plutonium-239, with a decay rate of 0.0028% per year:

• $T = \frac{\ln 2}{0.000028} \approx 24,755$ years

Thus, the half-life of plutonium-239 is about 24,755 years.

Check Your Understanding

• Cesium-137: $k = 0.023$ per year

• $T = \frac{\ln 2}{0.023} \approx 30.1$ years

• Barium-140: $T = 13$ days

• $k = \frac{\ln 2}{13} \approx 0.053 \approx 5.3\%$ per day

Applications in Life Science: Carbon Dating

Carbon-14 Dating

Carbon-14 has a half-life of 5730 years. The percentage of carbon-14 remaining in organic material can be used to estimate its age. The decay rate is:

• $k = \frac{\ln 2}{5730} \approx 0.00012097$, or $0.012097\%$ per year

Example: Determining Age Using Carbon Dating

If a linen wrapping from the Dead Sea Scrolls has lost 22.3% of its carbon-14, the remaining percentage is 77.7%:

• Set up the equation: $N = N_0 e^{-kt}$

• $0.777 N_0 = N_0 e^{-0.00012097 t}$

• $0.777 = e^{-0.00012097 t}$

• $\ln 0.777 = -0.00012097 t$

• $t = \frac{\ln 0.777}{-0.00012097} \approx 2086$ years

The linen is approximately 2086 years old.

Additional Example: Skeleton Age

If a skeleton has lost 60% of its carbon-14 (remaining 40%):

• $0.4 = e^{-0.00012097 t}$

• $t = \frac{\ln 0.4}{-0.00012097} \approx 7574$ years

The skeleton is approximately 7574 years old.

Applications in Business: Present Value

Present Value Formula

The present value ($P_0$) of an amount $P$ due $t$ years later at a continuous interest rate $k$ is:

• $P_0 = P e^{-kt}$

This formula is used to determine how much should be invested now to reach a desired future amount.

Example: Investment for Future Value

To grow $P_0$ to $10,000$ in 20 years at 4% interest compounded continuously:

• $10,000 = P_0 e^{0.04 \times 20}$

• $P_0 = \frac{10,000}{e^{0.8}} \approx 4,493.29$

At 6% interest:

• $P_0 = \frac{10,000}{e^{1.2}} \approx 3,011.94$

Summary Table: Key Formulas and Relationships

Concept	Formula	Description
Exponential Decay	$P(t) = P_0 e^{-kt}$	Amount remaining after time $t$
Half-life	$T = \frac{\ln 2}{k}$	Time to reduce to half the initial amount
Decay Rate	$k = \frac{\ln 2}{T}$	Rate of exponential decay
Present Value	$P_0 = P e^{-kt}$	Initial investment needed for future value $P$

Conclusion

Exponential and logarithmic functions are powerful tools in Business Calculus, enabling the modeling and analysis of decay, growth, and financial calculations. Understanding the relationships between decay rate, half-life, and present value is essential for solving applied problems in science and business.