Problem 85 - Calculus for Biology and Medicine

Question

Show that the function \(y=(1 / 2)^{x}\) can be written in the form \(y=e^{-\mu x}\), where \(\mu\) is a positive constant. Determine \(\mu .\)

Step-by-Step Solution

Answer

\( \mu = \ln 2 \)

Step 1: Understand the Problem

We need to express the function \( y=\left( \frac{1}{2} \right)^x \) in the form \( y=e^{-\mu x} \), where \( \mu \) is a positive constant. The goal is to determine the value of \( \mu \).

Step 2: Rewrite the Given Function Using Exponential Form

Express the base \( \frac{1}{2} \) as an exponential function. Remember \( a^x = e^{x \ln a} \). Therefore, set \( \left( \frac{1}{2} \right)^x = e^{x \ln \left( \frac{1}{2} \right)} \). This transforms the original expression using the natural exponential function.

Step 3: Simplify the Exponential Expression

We have \( e^{x \ln \left( \frac{1}{2} \right)} \). Rewrite this as \( e^{x (-\ln 2)} = e^{-x \ln 2} \). Now express it in the form \( e^{-\mu x} \) by identifying \( \mu \).

Step 4: Identify and Determine \( \mu \)

Compare \( e^{-x \ln 2} \) to \( e^{-\mu x} \) to identify \( \mu \). They match when \( \mu = \ln 2 \). Therefore, \( \mu \) is the natural logarithm of 2, \( \mu = \ln 2 \).

Step 5: Conclude the Solution

We have shown that \( y=\left( \frac{1}{2} \right)^x \) can be expressed as \( y = e^{-\mu x} \) with \( \mu = \ln 2 \), which confirms the problem's requirement.

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a logarithm with a base of \( e \). Here, \( e \) is a special constant approximately equal to 2.71828. This unique base makes the natural logarithm particularly useful in calculus and exponential growth models.

It offers a way to transform exponential expressions into more manageable forms, such as linear expressions. The key property that makes natural logarithms useful is: \( \ln(e^x) = x \). This property essentially "undoes" or reverses the exponential expression for an intuitive understanding.

• Used to solve equations involving exponentials.
• Helps in differentiating functions.
• Useful in expressing exponential growth or decay.

In the context of our exercise, we use the natural logarithm to transform the base \( \frac{1}{2} \) of the given function into a form that involves \( e \), facilitating simplification and analysis.

Exponential Form

Exponential form refers to expressing a number or variable in terms of an exponent. For example, \( a^x \) is an exponential expression where \( a \) is the base and \( x \) denotes the exponent. This is fundamental in mathematics, particularly in describing exponential growth or decay where values increase or decrease at rates proportional to their current value.

In our exercise, the function \( y = \left( \frac{1}{2} \right)^x \) needed to be written in a form that uses the natural number \( e \). Thanks to the property \( a^x = e^{x \ln a} \), we can express any exponential base in terms of \( e \).

Here’s how it works:

• Convert the base \( \frac{1}{2} \) using \( \ln \left( \frac{1}{2} \right) \).
• Introduce \( e \) by expressing it as \( e^{x \ln \left( \frac{1}{2} \right)} \).

This allows us to simplify and reframe exponential expressions to solve complex equations. Exponential functions are prevalent in real-world applications, including finance, science, and population studies.

Constant Determination

Finding a constant, like \( \mu \) in this problem, involves comparing known components of an expression to identify unambiguous values. Constants are integral as they define the specific characteristics of a mathematical model or relationship.

In our solution, we need to find \( \mu \) such that the function \( y = \left( \frac{1}{2} \right)^x \) can be equated to \( y = e^{-\mu x} \). By rewriting \( \left( \frac{1}{2} \right) \) in exponential terms, we found:

• \( e^{x \ln(\frac{1}{2})} = e^{-x \ln 2} \)
• Matching this against \( e^{-\mu x} \) gives us \( \mu = \ln 2 \).

By comparing these expressions, we identify the constant \( \mu \) directly as \( \ln 2 \), completing our solution.

Such techniques are vital in developing and understanding mathematical expressions and equations, ensuring accuracy and consistency in results.