Problem 56 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(\frac{119}{e^{6 x}-14}=7\)

Step-by-Step Solution

Answer

The approximate value of \(x\) is 0.567, to three decimal places.

Step 1: Simplification

Rearrange the equation to isolate the exponential term on one side. Multiply both sides by \((e^{6x} - 14)\) and divide by 7. This gives \(e^{6x} - 14 = \frac{119}{7}\).

Step 2: Solve for \(e^{6x}\)

Adding 14 to both sides of the equation, this yields: \(e^{6x} = 14 + \frac{119}{7} = \frac{217}{7}\).

Step 3: Take natural logarithm on both sides

Taking natural logarithms on both sides of the equation to get rid of the exponential gives: \(6x = \ln(\frac{217}{7})\).

Step 4: Solve for \(x\)

Divide both sides by 6 to solve for \(x\), this gives: \(x = \frac{\ln(\frac{217}{7})}{6}\). Now use a calculator to approximate the value to three decimal places.

Key Concepts

Natural Logarithm

The natural logarithm is an essential concept when dealing with exponential equations. It is denoted as \( \ln \) and is the inverse operation of taking an exponent with base \( e \), where \( e \approx 2.71828 \). Using the natural logarithm helps to solve equations involving the natural exponential function, \( e^x \), by transforming the problem into a form where algebraic manipulation can be used.

For example, if you have an equation \( e^y = a \), taking the natural logarithm on both sides gives \( y = \ln(a) \). This simplification is crucial because it changes an exponential equation into a linear one, making it easier to isolate the variable and solve for it.

In the context of the exercise, taking the natural logarithm on both sides of the equation \( e^{6x} = \frac{217}{7} \) allows us to bring down the exponent: \( 6x = \ln \left( \frac{217}{7} \right) \). This step is key in transitioning from the world of exponentials to that of arithmetic operations.

Algebraic Solutions

Algebraic solutions involve using a variety of algebraic manipulations to solve for the unknown variable. For exponential equations, this typically includes isolating the exponential term, taking logarithms, and performing basic arithmetic operations.

A step-by-step approach is often employed:

• Rearrange the equation to get the exponential term by itself.
• Use logarithms to remove the exponential, making the equation more manageable.
• Solve for the variable using arithmetic operations such as addition, subtraction, multiplication, or division.

In the given exercise, these steps are demonstrated as follows:

• Simplify the original equation, \( \frac{119}{e^{6x} - 14} = 7 \). This means isolating \( e^{6x} \) by rearranging and consolidating terms.
• Apply logarithms to transition from an exponential to a linear form.
• Solve for \( x \) using algebraic techniques, specifically division, to isolate \( x \).

This systematic approach helps in methodically breaking down complex problems into simpler, solvable steps.

Approximation Methods

Once an equation is simplified and solved algebraically, the result may need to be approximated. This is especially true when dealing with irrational numbers or when an exact decimal answer is not required.

Approximations are valuable because they provide a practical solution that can be easily interpreted and used for further calculations.

In the exercise, after finding the expression for \( x \), \( x = \frac{\ln(\frac{217}{7})}{6} \), a calculator is typically used to get a numerical approximation. This involves computing the natural logarithm and dividing it by 6 to get a decimal value. The problem specifically asks for this value to three decimal places for precision.

Using a calculator:

• Calculate \( \ln(\frac{217}{7}) \).
• Divide the result by 6.
• Round off the answer to three decimal places.

Approximation ensures that results are both understandable and meaningful, especially in real-world applications where exact numbers may not always be required.