Problem 13 - College Algebra

Question

A logarithmic model is given by the equation \(h(p)=67.682-5.792 \ln (p) .\) To the nearest hundredth, for what value of \(p\) does \(h(p)=62 ?\)

Step-by-Step Solution

Answer

The value of \( p \) is approximately 2.67.

Step 1: Understand the Equation

The equation representing the logarithmic model is given by \( h(p) = 67.682 - 5.792 \ln(p) \). We need to find the value of \( p \) such that \( h(p) = 62 \).

Step 2: Substitute and Rearrange

Substitute \( h(p) = 62 \) into the equation, yielding \( 62 = 67.682 - 5.792 \ln(p) \). Rearrange this to solve for \( \ln(p) \), which gives \( \ln(p) = \frac{67.682 - 62}{5.792} \).

Step 3: Calculate \( \ln(p) \)

Evaluate the expression \( \frac{67.682 - 62}{5.792} \) to find \( \ln(p) \). This simplifies to \( \ln(p) = \frac{5.682}{5.792} \approx 0.98096 \).

Step 4: Solve for \( p \)

To find \( p \), take the exponential of both sides of the equation \( \ln(p) = 0.98096 \). Thus, \( p = e^{0.98096} \).

Step 5: Compute \( p \)

Calculate \( p = e^{0.98096} \approx 2.666 \). Thus, rounding to the nearest hundredth, \( p \approx 2.67 \).

Key Concepts

Equations

An equation is a fundamental concept in mathematics that represents a statement in which two expressions are set equal to each other. In the context of logarithmic models, equations are used to describe relationships where one quantity depends on the logarithm of another.

Consider the given equation for the logarithmic model:

• \( h(p) = 67.682 - 5.792 \ln(p) \)

This is a specific type of equation where the variable \( p \) is related to the function \( h(p) \). It involves logarithms, which are inverse operations of exponential functions. Equations like this are utilized to model data that fits a logarithmic scale, where changes in one variable result in proportional changes in the logarithm of another variable.

Solving such equations requires isolating the variable of interest. In our example, the goal is to replace the function \( h(p) \) with a known value (62) to find the corresponding value of \( p \). This process involved:

• Substituting \( h(p) = 62 \) into the equation.
• Rearranging to solve for \( \ln(p) \).

Natural Logarithms

Natural logarithms use the base \( e \), where \( e \) is an irrational number approximately equal to 2.71828. They are denoted as \( \ln \) and are vastly used in mathematics and science due to their natural properties, such as dealing with growth or decay problems efficiently.

In the given exercise, a natural logarithm appears in the context of the function \( h(p) = 67.682 - 5.792 \ln(p) \). Here are some key points:

• \( \ln(p) \) denotes the natural logarithm of \( p \).
• It shows the relationship between \( p \) and the output \( h(p) \) in a logarithmic model.

The logarithmic nature means that changes in \( p \) don't change \( h(p) \) linearly but logarithmically. To solve for \( p \), one must first simplify the equation to isolate \( \ln(p) \). This is done by:

• Subtracting the constants from both sides.
• Dividing by the coefficient of \( \ln(p) \).

Once \( \ln(p) \) is isolated, it can be converted into \( p \) by using exponential functions, specifically the inverse operation of the logarithm.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are represented generally as \( y = a^x \), where \( a \) is the base and \( x \) is the exponent. These functions are key to solving equations involving logarithms, as they serve as the inverse operation.

In the context of the exercise, after we isolated \( \ln(p) \), which was calculated to be approximately 0.98096, the process to solve for \( p \) involved an exponential function:

• The equation \( \ln(p) = 0.98096 \) means \( p = e^{0.98096} \).

Here, the base \( e \) is used because we are working with natural logarithms. The transition from \( \ln(p) \) back to \( p \) uses the property that if \( \ln(x) = y \), then \( x = e^y \). This step is crucial as it effectively converts the logarithmic relationship back into its original form.Thus, exponential functions not only help solve for original values in logarithmic models but also illustrate the inverse relationship between logarithms and exponentials. This inverse relationship clarifies how exponential growth or decay can translate into linear relationships when viewed through a logarithmic lens.