Problem 26 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

In Problems 7 through 32, solve for \(x .\) $$ e^{x}-2=\frac{3}{e^{x}} $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = ln(3) \)

Step 1: Clear the Fraction

Multiply both sides of the equation by \(e^{x}\) to clear the fraction. This gives: \(e^{2x}-2e^{x}=3\)

Step 2: rearrange

Rearrange this equation so it looks like a quadratic equation in terms of \(e^{x}\), i.e., \(e^{2x}-2e^{x}-3=0\)

Step 3: Substitute

Let \(u = e^{x}\). Then the equation becomes a quadratic equation in terms of \(u\), i.e., \(u^{2}-2u-3=0\)

Step 4: Solve the Quadratic Equation

Factor the quadratic equation: \((u - 3)(u+1) = 0\). Therefore, \(u=3\) or \(u=-1\)

Step 5: Find \( x \)

Recall that \( u = e^{x} \). We find \( x \) by taking the natural log of both sides for each solution for \( u \). When \( u=3\), we find that \(x=ln(3)\). Because the base of the natural logarithm is \( e \) we cannot take the natural log of a negative number so we discard the solution \( u=-1\)

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation of degree two that can be written in the standard form:

• \( ax^2 + bx + c = 0 \)

Where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. The solutions to this quadratic equation are the values of \( x \) that make the equation true.
Quadratic equations can be solved using several different methods. These methods include:

• Factoring
• Completing the square
• Using the quadratic formula
• Graphing

In the provided exercise, the equation \( u^2 - 2u - 3 = 0 \) was solved by factoring it into \((u - 3)(u + 1) = 0 \). We then set each factor equal to zero to solve for \( u \).
Finally, remember that quadratic equations can have two real solutions, one real solution, or two complex solutions, depending on the value under the square root in the quadratic formula, known as the discriminant.

Natural Logarithms

Natural logarithms are a special type of logarithm where the base is an irrational constant known as "\( e \)", approximately equal to 2.71828.
The natural logarithm of a number \( x \) is written as \( \ln(x) \), and it tells us the power to which \( e \) must be raised to obtain \( x \).
The main properties of natural logarithms include:

• \( \ln(e) = 1 \)
• \( \ln(1) = 0 \)
• \( \ln(ab) = \ln(a) + \ln(b) \)
• \( \ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b) \)

In solving exponential equations like the one in the exercise, natural logarithms are essential for finding the variable in the exponent.
By applying the natural logarithm to both sides of equations such as \( e^x = 3 \), we can isolate \( x \) and solve it as \( x = \ln(3) \).
This use of natural logarithms provides a straightforward method to handle situations involving exponential growth and decay.

Exponential Functions

Exponential functions are mathematical expressions where a constant base is raised to a variable exponent. They have the general form:

• \( f(x) = a \cdot b^x \)

Where \( a \) is a constant coefficient, \( b \) is a positive real number (the base), and \( x \) is the exponent or variable.
In the context of the equation \( e^x - 2 = \frac{3}{e^x} \), the expression \( e^x \) represents the exponential function with base \( e \).
Exponential equations often arise in growth and decay problems such as population growth and radioactive decay.
To solve them, you may need to apply algebraic manipulation or connect the equation to logarithms, as demonstrated in the exercise.
The presence of \( e \) specifically relates to natural growth processes and has unique properties, such as a continuous rate of growth.
Understanding exponential functions helps explain situations where quantities increase or decrease at a rate proportional to their current value.