Problem 72 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the exponential equation algebraically. Round your result to three decimal places. Use a graphing utility to verify your answer. $$e^{2 x}-e^{x}-6=0$$

Step-by-Step Solution

Answer

The solution to the equation is \( x = \ln(3) \) or approximately 1.099 when rounded to three decimal places.

Step 1: Simplify the Equation

The expression \(e^{2 x}-e^{x}-6=0\) can be rewritten as a quadratic equation by letting \(y = e^{x}\). This substitution simplifies the equation to \(y^2 - y - 6 = 0\)

Step 2: Use the Quadratic Formula

Now, solve the equation \(y^2 - y - 6 = 0\) using the quadratic formula. The quadratic formula is \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Hence, substituting \(a = 1\), \(b = -1\), and \(c = -6\) into the quadratic formula gives two solutions for \(y\): \(y_1 = 3\) and \(y_2 = -2\)

Step 3: Substitute Back

Next, substitute \(y = e^{x}\) back into the equation to solve for \(x\). Therefore, \(e^{x} = 3\) and \(e^{x} = -2\). The second equation has no solution since the base of natural logarithm \(e^{x}\) can never be negative. For the first equation, using the definition of natural logarithm, we get an answer \(x = \ln(3)\) which rounds to 1.099.

Step 4: Verify Using a Graphing Utility

To confirm the solution, graph \(y = e^{2 x}\), \(y = e^{x}\), and \(y = 6\) using a graphing calculator or software. The intersection points of the graphs should verify the solution.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). It provides us a straightforward way to find the roots of any quadratic equation, regardless of the method needed to factor it. The formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation. To solve using this formula, you first need to calculate the discriminant \( b^2 - 4ac \), which determines the nature of the roots. If the discriminant is positive, there are two real and distinct solutions. A discriminant of zero results in one real, repeated root, and a negative discriminant indicates complex roots. In our specific exercise, by substituting the given values (\( a = 1 \), \( b = -1 \), \( c = -6 \)), we found two solutions for \( y \): 3 and -2.

Natural Logarithm

Natural logarithms are logarithms with the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828. They are denoted as \( \ln(x) \). Natural logarithms are particularly useful in solving equations involving exponential terms because they "undo" the exponential function.
For an equation \( e^x = y \), taking the natural logarithm allows you to isolate the variable \( x \): \( x = \ln(y) \).
In our problem, after simplifying the equation to solve for \( y = e^x \), we had \( e^x = 3 \). By applying the natural logarithm, the solution for \( x \) becomes \( x = \ln(3) \). This process is vital as it translates exponential expressions into linear forms easy to handle algebraically. It’s important to note that \( e^x = -2 \) has no solution in real numbers because exponential functions never yield negative outputs.

Graphing Utility

Graphing utilities, such as graphing calculators or software like Desmos and GeoGebra, are powerful tools to visualize equations and verify solutions. They serve multiple purposes, including:

• Graphing the individual functions involved in the equation.
• Identifying points of intersection which represent solutions to the equations.

In this exercise, to check our algebraic solution, we graph \( y = e^{2x} \), \( y = e^x \), and \( y = 6 \) as separate functions. The intersections of these graphs confirm the solutions we derived algebraically. If the graph shows an intersection point at, say, \( x = 1.099 \), it verifies our algebraically calculated result \( x = \ln(3) \), making sure we've done everything correctly.

Algebraic Solution

An algebraic solution involves solving an equation using algebraic manipulations and rules. This means using foundational operations such as simplifying, factoring, or performing arithmetic operations to isolate the variable and find its value. In this case, the given exponential equation was first transformed into a quadratic one by substituting \( y = e^x \), leading us to the equation \( y^2 - y - 6 = 0 \).

• This transformation is key to making the equation solvable through the quadratic formula.
• Once \( y \) is found, it’s substituted back, leading to an easier exponential equation.

The algebraic approach is systematic, ensuring accuracy and completeness in finding solutions. It also highlights the importance of substitution in solving complex equations by temporarily altering the form.

Intersection Points

Intersection points in the context of graphing functions signify the solutions to equations formulated in context. When you graph two or more functions, the points at which they cross are their intersection points. These points are critical in verifying solutions from algebraic equations.

• If the functions are correctly graphed and the calculated points coincide with these intersections, the solutions are confirmed as correct.
• In our particular exercise, the graph uses equations like \( y = e^{2x} \), \( y = e^x \), and \( y = 6 \).

Finding intersection points helps show where the input values meet specific conditions in different functions. Thus, if the intersection is at \( x \approx 1.099 \), it demonstrates our algebraic solution's validity with \( x = \ln(3) \). Not just reinforcing it but also providing a visual confirmation of the work done.