Exponential and Logarithmic Functions of the Natural Base $e$

Exponential and Logarithmic Functions of the Natural Base, e

Introduction

This section explores exponential and logarithmic functions with the natural base e, a fundamental concept in Business Calculus. These functions are widely used in modeling continuous growth and decay, such as in finance, population studies, and natural processes.

Definition of the Natural Base e

Understanding e

• Natural base, denoted e, is an irrational constant approximately equal to 2.71828.

• It is defined by the limit: $e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$

• e arises naturally in problems involving continuous growth or decay.

Continuous Exponential Growth

Theorem 1: Continuous Exponential Growth

• If a quantity P grows continuously at an annual percentage rate r (expressed as a decimal), its future value after t years is given by: $A = Pe^{rt}$

• This formula is essential for modeling investments, populations, and other continuously changing quantities.

Example: Continuous Growth

• Problem: Luis invests $5000 in an account earning 3.25% interest compounded continuously for 5 years. Find the future value.

• Solution:

  • $P = 5000$, $r = 0.0325$, $t = 5$

  • $A = 5000e^{0.0325 \times 5}$

  • $A \approx 5882.24$

Example: Practice Problem

• Problem: Nari invests $20,000 at 4% interest compounded continuously for 3 years. Find the future value.

• Solution:

  • $P = 20,000$, $r = 0.04$, $t = 3$

  • $A = 20,000e^{0.04 \times 3}$

  • $A \approx 22,549.94$

Natural Logarithms

Definition: Natural Logarithm

• The natural logarithm of a positive number x is the logarithm with base e, denoted $\ln x$.

• $\ln x = \log_e x$

• The equation $y = \ln x$ is equivalent to $e^y = x$.

Example: Evaluating Natural Logarithms

• Find $\ln e^4$, $\ln 1$, $\ln e^{-1}$, and $\ln 20$.

• Solutions:

  • $\ln e^4 = 4$ (since $e^4$ is the inverse of $\ln$)

  • $\ln 1 = 0$ (since $e^0 = 1$)

  • $\ln e^{-1} = -1$ (since $e^{-1}$ is the inverse of $\ln$)

  • $\ln 20 \approx 2.996$ (using a calculator)

Properties of Natural Logarithms

Theorem 2: Properties

• $\ln(MN) = \ln M + \ln N$

• $\ln\left(\frac{M}{N}\right) = \ln M - \ln N$

• $\ln(M^k) = k \ln M$

• $\ln e = 1$

• $\ln 1 = 0$

• Change of base formulas:

  • $\log_b M = \frac{\ln M}{\ln b}$

  • $\ln M = \frac{\log M}{\log e}$

• $e^{\ln x} = x$, for all $x > 0$

• $\ln(e^x) = x$, for all real $x$

Example: Applying Properties

• Given $\ln 2 = 0.6931$ and $\ln 3 = 1.0986$, find:

• $\ln 6 = \ln(2 \cdot 3) = \ln 2 + \ln 3 = 0.6931 + 1.0986 = 1.7917$

• $\ln 81 = \ln(3^4) = 4 \ln 3 = 4 \times 1.0986 = 4.3944$

• $\ln \left(\frac{1}{3}\right) = \ln 1 - \ln 3 = 0 - 1.0986 = -1.0986$

• $\ln(2e^5) = \ln 2 + \ln(e^5) = 0.6931 + 5 = 5.6931$

• $\ln 3 \approx 1.0986$ (as given)

Summary Table: Properties of Natural Logarithms

Property	Formula	Description
Product	$\ln(MN) = \ln M + \ln N$	Logarithm of a product is the sum of logarithms
Quotient	$\ln\left(\frac{M}{N}\right) = \ln M - \ln N$	Logarithm of a quotient is the difference of logarithms
Power	$\ln(M^k) = k \ln M$	Logarithm of a power is the exponent times the logarithm
Identity	$\ln e = 1$	Natural log of e is 1
Zero	$\ln 1 = 0$	Natural log of 1 is 0
Inverse	$e^{\ln x} = x$	Exponential and logarithm are inverse functions
Inverse	$\ln(e^x) = x$	Logarithm and exponential are inverse functions

Domain and Range of Natural Logarithm Functions

Domain and Range

• The function $f(x) = \ln x$ is defined only for $x > 0$.

• Domain: $(0, \infty)$

• Range: $(-\infty, \infty)$ (all real numbers)

• The function is always increasing.

Example: Finding the Domain

• Find the domain of $f(x) = \ln(5 - 2x)$.

• Set $5 - 2x > 0 \implies x < 2.5$

• Domain: $(-\infty, 2.5)$

Solving Exponential and Logarithmic Equations

Solving Exponential Equations Using Logarithms

• To solve equations of the form $e^{kx} = a$, take the natural logarithm of both sides.

• Example: Solve $e^{3x} = 2$

• Take $\ln$ of both sides: $3x = \ln 2$

• $x = \frac{\ln 2}{3} \approx 0.231$

Solving for Time in Continuous Growth

• Example: $500e^{0.015t} = 750$

• Divide both sides by 500: $e^{0.015t} = 1.5$

• Take $\ln$ of both sides: $0.015t = \ln 1.5$

• $t = \frac{\ln 1.5}{0.015} \approx 27.03$

• Note: The original notes had $e^{0.015t} = 3$, so $t = \frac{\ln 3}{0.015} \approx 73.24$

Exponential Growth Rate and Doubling Time

Theorem 4: Relationship Between Growth Rate and Doubling Time

• The exponential growth rate $r$ and the doubling time $T$ are related by: $r = \frac{\ln 2}{T}$ $T = \frac{\ln 2}{r}$

• This relationship is useful for quickly estimating how long it takes for a quantity to double at a given growth rate.

Example: Business Application

• Suppose a messaging app doubles its users every 8 months. What is the exponential growth rate?

• $r = \frac{\ln 2}{8} \approx 0.087$ per month

Solving Logarithmic Equations

Example: Solving for x

• Solve $5 + 7\ln(x + 3) = 19$

• Isolate the logarithmic term: $7\ln(x + 3) = 14$

• $\ln(x + 3) = 2$

• $x + 3 = e^2$

• $x = e^2 - 3 \approx 4.389$