Problem 84 - A Graphical Approach to Precalculus with Limits

Question

The given equations are quadratic in form. Solve each and give exact solutions. $$3 e^{2 x}+2 e^{x}=1$$

Step-by-Step Solution

Answer

The exact solution is \( x = -\ln(3) \).

Step 1: Substitute for Simplicity

Let \( u = e^x \). Then \( e^{2x} = (e^x)^2 = u^2 \). Substitute these into the equation to obtain \( 3u^2 + 2u - 1 = 0 \).

Step 2: Recognize the Quadratic Equation

The equation \( 3u^2 + 2u - 1 = 0 \) is a quadratic equation in terms of \( u \). It is in the standard form \( ax^2 + bx + c = 0 \) where \( a = 3 \), \( b = 2 \), and \( c = -1 \).

Step 3: Use the Quadratic Formula

The quadratic formula provides solutions for \( ax^2 + bx + c = 0 \) as \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 3 \), \( b = 2 \), and \( c = -1 \) into the formula to get \( u = \frac{-2 \pm \sqrt{2^2 - 4 \times 3 \times (-1)}}{2 \times 3} \).

Step 4: Simplify Under the Square Root

Calculate the discriminant \( b^2 - 4ac = 2^2 - 4 \times 3 \times (-1) = 4 + 12 = 16 \). So, the square root is \( \sqrt{16} = 4 \).

Step 5: Solve for u

Substitute the square root back into the quadratic formula: \( u = \frac{-2 \pm 4}{6} \). This gives two solutions: \( u = \frac{2}{6} = \frac{1}{3} \) and \( u = \frac{-6}{6} = -1 \).

Step 6: Convert Back to e^x and Solve for x

Since \( u = e^x \), we have \( e^x = \frac{1}{3} \) and \( e^x = -1 \). The solution for \( e^x = -1 \) does not exist for real numbers. So solve \( e^x = \frac{1}{3} \) to get \( x = \ln \left( \frac{1}{3} \right) = -\ln(3) \).

Key Concepts

Exponential Functions

Exponential functions are a common type of mathematical expression where a variable represents the exponent of a constant base. These functions are often of the form \( f(x) = a^x \), where \( a \) is a positive constant and \( x \) is any real number. In our exercise, the exponential function \( e^{x} \) was utilized, where \( e \) is an irrational number approximately equal to 2.718. The reason why exponential functions are significant is because they model growth and decay processes.

In the equation \( 3e^{2x} + 2e^x = 1 \), we see how an exponential function can represent an unknown variable \( x \) in its exponent. To simplify such equations, we can perform a substitution trick, as shown with \( u = e^x \) in this problem.

• Substitution transforms the equation into a quadratic form, making it more manageable.
• By rewriting \( e^{2x} \) as \( u^2 \), the equation takes a recognizable quadratic form: \( 3u^2 + 2u - 1 = 0 \).

Exponential functions are powerful tools in various fields, including finance, biology, and physics, because of their ability to describe diverse natural phenomena.

Quadratic Formula

The quadratic formula is a mathematical solution technique designed for solving quadratic equations, equations of the form \( ax^2 + bx + c = 0 \). In our example, this formula was essential in finding the solutions of \( 3u^2 + 2u - 1 = 0 \). The formula provides the roots of a quadratic equation, and is expressed as:\[ u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This solution method is particularly useful because it applies to any quadratic equation, offering a systematic approach to calculate the roots. In our problem:

• The coefficients are \( a = 3 \), \( b = 2 \), and \( c = -1 \).
• Substitute these values into the formula to account for the terms under the square root and solve the equation.
• The discriminant \( b^2 - 4ac \) helps determine the nature of the roots:
• If positive, two distinct real solutions exist.
• If zero, one real solution (a double root) is present.
• If negative, there are no real solutions, only complex ones.

For our equation, the discriminant was 16, allowing us to calculate two real solutions for \( u \): \( \frac{1}{3} \) and \( -1 \). Utilizing the quadratic formula simplifies the process of solving equations and ensures accuracy.

Logarithms

Logarithms are the inverses of exponential functions and are essential when solving equations where the variable is an exponent. In our exercise, after transforming the quadratic equation back to an exponential equation \( u = e^x \), we encountered \( e^x = \frac{1}{3} \). To solve for \( x \), we apply logarithms.

Using the natural logarithm (\( \ln \)), which corresponds to the base \( e \), helps us determine:

• When \( e^x = \frac{1}{3} \), \( x = \ln\left(\frac{1}{3}\right) \), simplifying it to \( -\ln(3) \).
• The natural logarithm allows for determining the value of \( x \) by 'bringing down' the exponent.

This illustrates the concept:

• \( \ln \) is frequently used in calculations involving rates and decay because it directly relates to the number \( e \).
• Logarithms convert multiplicative relationships into additive ones, which forms the foundation of logarithmic scales.

Logarithms empower us to solve complex exponential equations, enabling various practical applications from computing time in decay processes to complex sound level scales in acoustics.