Problem 438 - Precalculus

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 2.66.

Step 1: Substitute the value into the equation

Given the equation \( h(p) = 67.682 - 5.792 \ln(p) \), we know \( h(p) \) is 62. Substitute this value into the equation: \[ 62 = 67.682 - 5.792 \ln(p) \].

Step 2: Rearrange the equation

To isolate the logarithmic term, subtract 67.682 from both sides: \[ 62 - 67.682 = -5.792 \ln(p) \]. This simplifies to \[ -5.682 = -5.792 \ln(p) \].

Step 3: Solve for the logarithm

Divide both sides by -5.792 to solve for \( \ln(p) \): \[ \ln(p) = \frac{-5.682}{-5.792} \]. Calculating the right-hand side gives \( \ln(p) \approx 0.9801 \).

Step 4: Exponentiate to solve for p

To solve for \( p \), exponentiate both sides: \( e^{\ln(p)} = e^{0.9801} \). This simplifies to \( p \approx e^{0.9801} \).

Step 5: Calculate the final value of p

Use a calculator to find \( e^{0.9801} \approx 2.6647 \). To the nearest hundredth, this gives \( p = 2.66 \).

Key Concepts

Logarithmic Functions

Logarithmic functions are mathematical functions that help us understand how quantities change multiplicatively. They are inverses of exponential functions, meaning while an exponential function grows very quickly, a logarithmic function tends to increase more gradually. This slow increase makes them ideal for dealing with scales of multiplication and division rather than addition and subtraction.
For example, in the equation \( h(p) = 67.682 - 5.792 \ln(p) \), the part \( \ln(p) \) is the natural logarithm of \( p \). The natural logarithm has a special base known as Euler's number \( e \), approximately equal to 2.718.
Logarithms can simplify multiplication problems into manageable addition problems. For students who might find this challenging, remember:

• Logarithms help in simplifying complex equations.
• The natural logarithm \( \ln(x) \) implies the power to which \( e \) must be raised to obtain \( x \).
• Logarithmic scales are useful for showing data that changes exponentially or has huge variance in numbers.

Exponential Functions

Exponential functions describe processes of growth or decay that are fast and sustained over time. They are the counterparts to logarithmic functions. In an exponential function of form \( f(x) = a \cdot b^x \), \( a \) represents the initial amount, and \( b \) tells us how much it multiplies by each step or unit of \( x \).
In the solution, after finding \( \ln(p) \), we used an exponential function to solve for \( p \) by taking \( e^{0.9801} \). This means we exponentiate 0.9801 to find the original value \( p \).
Some key points to remember about exponential functions are:

• They exhibit rapid changes of values, either growth or decay.
• \( e^{x} \) is exponential growth where \( e \) is the base of the natural logarithm.
• Exponential functions appear frequently in natural processes, such as population growth or radioactive decay.

Solving Equations

Solving logarithmic or exponential equations often requires several strategic steps. The given exercise runs through common methods employed in addressing such equations, including rearrangement, proportional reasoning, and exponentiation.
Starting from the equation \( 62 = 67.682 - 5.792 \ln(p) \), the primary challenge involved isolating \( \ln(p) \) to eventually solve for \( p \). To do this, we followed these key steps:

• Substitute and rearrange: This involves substituting given values and rearranging the equation to isolate terms associated with the variable of interest.
• Use operations to simplify: Such as dividing both sides of the equation to make it easier to isolate the logarithmic term.
• Exponentiate to solve for the variable: This takes advantage of knowing that taking the exponential of a log essentially "undoes" the log, hence finding the value of the variable.
• Calculate and approximate: Once isolated, we use calculators or computational tools to find numerical approximations to high precision.

When dealing with logarithmic or exponential equations, breaking down each step and checking your work carefully will simplify what initially seems complicated.