Question

Solve the exponential equation algebraically. Approximate the result to three decimal places.\(e^{2 x}-3 e^{x}-4=0\)

Short answer

The approximate solution to the equation \(e^{2 x}-3 e^{x}-4=0\) is \(x = 1.386\)

Step-by-step solution

Transformation

The equation \(e^{2 x}-3 e^{x}-4=0\) can be arranged in quadratic form if we let \(y = e^{x}\). This simplifies the equation to: \(y^2 - 3y - 4 = 0\).

Solving the quadratic equation

We can solve the quadratic equation by using the quadratic formula \(y = [-b ± sqrt(b^2 - 4ac)] / (2a)\). Applying this formula, we find that \(y1 = 4\) and \(y2 = -1\).

Back-substitution

Now that we have our 'y' values, we substitute them back into the original substitution \(y = e^{x}\), to find x. This gives us: \(x = ln(y)\), and with \(y1 = 4\) and \(y2= -1\), we get two cases. Remembering that the logarithm of any negative number is undefined, only \(x = ln(4)\) gives us meaningful result.

Calculating result

We now calculate the value of \(x = ln(4)\), which is approximately \(x = 1.386\), when rounded to three decimal places.

Key concepts

Algebraic Solving Methods

When facing exponential equations, algebraic solving methods are a great tool, as they provide a structured way to isolate the variable of interest. The key is to transform or simplify complex equations into more manageable forms. Here's how you can approach it:

• Identifying the structure: Recognize patterns that would help reduce the equation into a simpler form. Look for ways to convert exponential terms into algebraic ones using substitutions.


• Substitution Method: For the equation \(e^{2x} - 3e^x - 4 = 0\), we used substitution. By letting \(y = e^x\), the equation turns into a quadratic equation, \(y^2 - 3y - 4 = 0\). This is a powerful way to handle complex exponents.


• Solving the simplified equation: Once transformed, the problem becomes one of solving a familiar quadratic equation. This method not only simplifies things but also makes it easier to retrieve the original variable by back substitution.

Understanding these methods will enable you to confidently solve exponential equations consistently and accurately.

Quadratic Equations

Quadratic equations frequently appear when solving algebraic problems that involve square terms. A quadratic equation has the form \(ax^2 + bx + c = 0\), and can be solved using various methods:

• Quadratic Formula: This formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) allows us to solve any quadratic equation by simply substituting the coefficients \(a\), \(b\), and \(c\).


• Factoring: This involves expressing the quadratic equation as a product of its roots. It is efficient but not always applicable, especially if the roots aren't integers or rational numbers.


• Completing the Square: This method reconfigures the equation to make the variable easier to isolate. Although more involved, this technique can offer insights into the properties of quadratic equations.

In the given exercise, the quadratic formula was the method used because it provides a straightforward calculation even when the roots involve irrational numbers or don't factor neatly. Being familiar with these methods will allow you to tackle any quadratic equation with ease.

Natural Logarithms

Natural logarithms, denoted as \(\ln\), are essential in dealing with exponential equations, especially when trying to solve for unknown exponents. They are the inverse of the exponential function with base \(e\), the natural constant, approximately equal to 2.71828. Here's how they work:

• Properties of Natural Logarithms: Understanding its properties, such as \(\ln(e^x) = x\) and \(e^{\ln(x)} = x\), is crucial. These properties allow us to transition seamlessly between exponential and logarithmic forms.


• Using \(\ln\) in Solving Equations: In the exercise, after finding the possible values of \(y\) from the quadratic equation, back-substitution with the formula \(x = \ln(y)\) was used to find \(x\). For example, \(x = \ln(4)\) calculated to approximately 1.386.


• Handling Undefined Logarithms: It's important to remember that \(\ln\) of a negative or zero value is undefined, which was why \(y = -1\) was disregarded in the solution.

Natural logarithms provide a very straightforward and rational way to solve equations involving exponential terms, converting them into analyzable forms.