Exponential And Logarithmic Functions Examples With Solutions

Exponential and Logarithmic Functions: Examples with Solutions

Exponential and logarithmic functions are fundamental concepts in mathematics with widespread applications in various fields, including science, engineering, finance, and computer science. Understanding these functions, their properties, and how to solve problems involving them is crucial for success in many academic and professional pursuits. This comprehensive guide delves into the intricacies of exponential and logarithmic functions, providing numerous examples with detailed solutions to solidify your understanding.

Understanding Exponential Functions

An exponential function is a function of the form f(x) = a^x, where 'a' is a positive constant called the base, and 'x' is the exponent, which can be any real number. The key characteristic of an exponential function is that the independent variable ('x') appears as the exponent.

Key Properties of Exponential Functions:

• Growth/Decay: If a > 1, the function represents exponential growth; if 0 < a < 1, it represents exponential decay.
• Horizontal Asymptote: The x-axis (y = 0) serves as a horizontal asymptote for exponential functions. The graph approaches but never touches this line.
• Domain and Range: The domain of an exponential function is all real numbers (-∞, ∞), while the range is (0, ∞) (all positive real numbers).
• One-to-one Function: Each input value ('x') corresponds to a unique output value ('y'), and vice versa. This property is essential for the inverse relationship with logarithmic functions.

Example 1: Exponential Growth

A population of bacteria doubles every hour. If the initial population is 100, find the population after 5 hours.

Solution:

The formula for exponential growth is given by: P(t) = P₀ * a^t, where P(t) is the population at time t, P₀ is the initial population, a is the growth factor, and t is the time.

In this case, P₀ = 100, a = 2 (doubles every hour), and t = 5 hours.

P(5) = 100 * 2⁵ = 100 * 32 = 3200

Therefore, the population after 5 hours will be 3200 bacteria.

Example 2: Exponential Decay

The half-life of a radioactive substance is 10 years. If you start with 50 grams, how much will remain after 30 years?

Solution:

The formula for exponential decay is: A(t) = A₀ * (1/2)^(t/h), where A(t) is the amount remaining at time t, A₀ is the initial amount, h is the half-life, and t is the time.

Here, A₀ = 50 grams, h = 10 years, and t = 30 years.

A(30) = 50 * (1/2)^(30/10) = 50 * (1/2)³ = 50 * (1/8) = 6.25 grams

After 30 years, 6.25 grams of the substance will remain.

Understanding Logarithmic Functions

A logarithmic function is the inverse of an exponential function. If y = a^x, then the logarithmic function is written as x = log_ay. This is read as "x is the logarithm of y to the base a." In simpler terms, the logarithm answers the question: "To what power must we raise the base 'a' to get y?"

Key Properties of Logarithmic Functions:

• Inverse Relationship: Logarithmic functions and exponential functions are inverses of each other. This means that f(f^-1(x)) = x and f^-1(f(x)) = x, where f(x) is the exponential function and f^-1(x) is its logarithmic inverse.
• Domain and Range: The domain of a logarithmic function is (0, ∞) (all positive real numbers), and the range is (-∞, ∞) (all real numbers).
• Vertical Asymptote: The y-axis (x = 0) acts as a vertical asymptote; the graph approaches but never touches this line.
• Logarithm Rules: Several crucial rules simplify logarithmic calculations (see below).

Common Logarithms (base 10) and Natural Logarithms (base e)

• Common Logarithms: These use base 10 and are often written as log(x) or log₁₀(x).
• Natural Logarithms: These use the base e (Euler's number, approximately 2.718), and are written as ln(x) or log_e(x).

Logarithm Rules:

• Product Rule: log_a(xy) = log_a(x) + log_a(y)
• Quotient Rule: log_a(x/y) = log_a(x) - log_a(y)
• Power Rule: log_a(xⁿ) = n * log_a(x)
• Change of Base Rule: log_a(x) = log_b(x) / log_b(a) (This is useful for changing the base to a more convenient one, such as 10 or e for calculator use.)

Example 3: Solving a Logarithmic Equation

Solve for x: log₂(x) + log₂(x - 2) = 3

Solution:

Using the product rule, we combine the logarithms:

log₂(x(x - 2)) = 3

Rewrite in exponential form:

x(x - 2) = 2³

x² - 2x = 8

x² - 2x - 8 = 0

Factor the quadratic equation:

(x - 4)(x + 2) = 0

x = 4 or x = -2

However, since the logarithm is undefined for negative numbers, we discard x = -2. Therefore, the solution is x = 4.

Example 4: Using Logarithms to Solve an Exponential Equation

Solve for x: 3^x = 10

Solution:

Take the logarithm of both sides (using either base 10 or base e):

log(3^x) = log(10)

x * log(3) = 1

x = 1 / log(3)

Using a calculator, x ≈ 2.096

Example 5: Applications in Finance – Compound Interest

The formula for compound interest is A = P(1 + r/n)^(nt), where:

• A = the future value of the investment/loan, including interest
• P = the principal investment amount (the initial deposit or loan amount)
• r = the annual interest rate (decimal)
• n = the number of times that interest is compounded per year
• t = the number of years the money is invested or borrowed for

Suppose you invest $1000 at an annual interest rate of 5%, compounded monthly. How much money will you have after 10 years?

Solution:

P = 1000, r = 0.05, n = 12 (monthly compounding), t = 10

A = 1000(1 + 0.05/12)^(12*10) = 1000(1 + 0.004167)¹²⁰ ≈ 1647.01

After 10 years, you will have approximately $1647.01.

Example 6: Applications in Science – Radioactive Decay

We revisited radioactive decay earlier. Let's explore a slightly different scenario. A radioactive substance decays according to the equation A(t) = A₀e^-kt, where A(t) is the amount remaining at time t, A₀ is the initial amount, k is the decay constant, and t is the time. If the half-life is 5730 years (Carbon-14), find the decay constant k.

Solution:

When t = 5730 years, A(t) = A₀/2 (half-life). Substitute these values into the equation:

A₀/2 = A₀e^-k*5730

Divide both sides by A₀:

1/2 = e^-k*5730

Take the natural logarithm of both sides:

ln(1/2) = -k * 5730

k = -ln(1/2) / 5730 ≈ 0.00012097

The decay constant k is approximately 0.00012097 per year.

Solving More Complex Problems Involving Exponential and Logarithmic Functions

Many real-world problems require the combined application of exponential and logarithmic functions. Solving these often involves manipulating logarithmic properties, employing change-of-base rules, and using algebraic techniques. Let's consider an example:

Example 7: A Combined Problem

The number of subscribers to a social media platform grows exponentially according to the function N(t) = 1000e^0.15t, where N(t) is the number of subscribers in thousands and t is the time in years since the platform launched. When will the number of subscribers reach 100,000?

Solution:

We set N(t) = 100 (since N(t) is in thousands), and solve for t:

100 = 1000e^0.15t

Divide by 1000:

0.1 = e^0.15t

Take the natural logarithm of both sides:

ln(0.1) = 0.15t

t = ln(0.1) / 0.15 ≈ -15.35 years.

The negative sign indicates that the number of subscribers was already at 100,000 before the platform launched. This might indicate an error in the model, suggesting that the initial exponential growth assumption might not hold true for the early life of the platform. A more accurate model might be needed, perhaps incorporating a different growth function for the initial phase.

This example showcases how careful analysis is necessary when applying mathematical models to real-world scenarios. The result might require a reevaluation of the underlying assumptions and adjustments to the model for a more accurate prediction.

This comprehensive guide provides a solid foundation in understanding and working with exponential and logarithmic functions. By mastering these concepts and applying the examples and solutions provided, you'll be well-equipped to tackle numerous challenges across various disciplines. Remember to practice consistently and explore additional resources to further deepen your understanding.