Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions

Problem 7.2.54

Chapter 7, Problem 7.2.54

Geometric means A quantity grows exponentially according to y(t) = y₀eᵏᵗ. What is the relationship among m, n, and p such that y(p) = √(y(m)y(n))?

Verified step by step guidance

Start with the given exponential growth function: \(y(t) = y_0 e^{k t}\), where \(y_0\) and \(k\) are constants.

Write the expressions for \(y(m)\), \(y(n)\), and \(y(p)\) using the function: \(y(m) = y_0 e^{k m}\), \(y(n) = y_0 e^{k n}\), and \(y(p) = y_0 e^{k p}\).

Set up the equation given by the problem: \(y(p) = \sqrt{y(m) y(n)}\).

Substitute the expressions for \(y(m)\), \(y(n)\), and \(y(p)\) into the equation: \(y_0 e^{k p} = \sqrt{(y_0 e^{k m})(y_0 e^{k n})}\).

Simplify the right side by multiplying inside the square root and then taking the square root, and then solve for the relationship among \(m\), \(n\), and \(p\) by equating the exponents.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Exponential Growth Function

An exponential growth function is expressed as y(t) = y₀e^{kt}, where y₀ is the initial value, k is the growth rate, and t is time. The function models quantities that increase proportionally to their current value, leading to rapid growth over time.

Geometric Mean

The geometric mean of two positive numbers a and b is √(ab). It represents the central tendency by multiplying the numbers and taking the square root, often used to find a value that balances exponential relationships between quantities.

Properties of Exponents and Logarithms

Exponential expressions like e^{kt} follow rules such as e^{a} * e^{b} = e^{a+b}. Taking logarithms can simplify equations involving exponentials, allowing us to solve for variables by converting multiplicative relationships into additive ones.

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