Problem 45 - College Algebra and Calculus: An Applied Approach

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(7-2 e^{x}=6\)

Step-by-Step Solution

Answer

The solution to the equation is \(x \approx -0.693\)

Step 1: Isolate the term with exponent

To isolate exponent term \(2 e^{x}\), subtract 6 from both sides of the equation \(7-2 e^{x}=6\), which gives: \(2 e^{x}=7-6 = 1\)

Step 2: Eliminate 2 from the left side of equation

Divide both sides of the equation \(2 e^{x}=1\) by 2, resulting in: \(e^{x}=0.5\)

Step 3: Apply natural logarithm to both sides

Applying a natural logarithm to both sides of the equation \(e^{x}=0.5\), we get: \(x = ln(0.5)\)

Step 4: Approximate the result

Using a calculator to evaluate \(ln(0.5)\), approximate to three decimal places: \(x \approx -0.693 \)

Key Concepts

Algebraic Methods in Calculus

Algebraic methods are at the core of solving calculus problems, particularly when dealing with limits, derivatives, and integrals. When faced with an exponential equation, such as the textbook exercise (7 - 2e^{x} = 6), algebra can help isolate the variable of interest before applying calculus techniques.

Firstly, we aim to get the term containing the variable alone, which typically involves basic operations such as addition, subtraction, multiplication, or division. In our exercise, we subtracted 6 from both sides to isolate (2e^{x}) and then divided by 2 to solve for (e^{x}). These algebraic manipulations pave the way for applying more advanced calculus concepts, such as taking logarithms or derivatives, to find the value of (x).

• Identify and isolate terms with the variable
• Use arithmetic operations to solve for the variable
• Prepare the equation for calculus techniques (logarithms, derivatives...)

Natural Logarithms

Natural logarithms are a cornerstone of solving exponential equations, allowing us to 'unlock' the variable from the exponent. In the context of the given problem, once we've isolated (e^{x}) and equated it to 0.5, applying the natural logarithm to both sides is the next logical step. The natural logarithm is the inverse operation of raising (e) to a power, which means that (ln(e^{x}) = x).

By applying the natural logarithm to both sides of the equation, (ln(e^{x}) = ln(0.5)), we directly find the value of (x). This is because (ln(e^{x})) simplifies to (x), resulting in the equation (x = ln(0.5)). Here, the natural logarithm serves as an essential tool for solving exponential equations when the base is (e).

• The natural logarithm is the inverse of the exponential function with base (e)
• Applying (ln) to both sides of an equation can reveal the exponent
• It's essential when dealing with exponential growth and decay problems

Exponential Equations Approximation

In scenarios where exact expressions are less practical or unnecessary, we approximate the values of exponential equations to a desired precision. After taking the natural logarithm and arriving at the equation (x = ln(0.5)), our final step involves approximating the result. Since ln(0.5) yields an irrational number, we round it to a fixed number of decimal places for easier interpretation and practicality.

Using a calculator or computational software, we find that (ln(0.5)) is approximately -0.693 to three decimal places. This level of precision is often sufficient in applied contexts, such as finance, physics, or engineering, where real-world measurements are inherently approximate. Approximation also simplifies the communication of results, making them more digestible.

• Rounding irrational numbers to a fixed decimal place makes them usable
• Approximation aligns mathematical results with real-world application
• It facilitates easier comparison and discussion of computed values