Problem 84 - A Graphical Approach to College Algebra

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the substitution

The equation is given as \(3 e^{2x} + 2 e^{x} = 1\). The substitution \(v = e^{x}\) can be used because \(e^{2x} = (e^x)^2 = v^2\). Substitute this expression to transform the equation.

Step 2: Transform the equation

Substitute \(v = e^{x}\) into the equation. Thus, \(3(e^{x})^2 + 2e^{x} = 1\) transforms to \(3v^2 + 2v = 1\).

Step 3: Rearrange into standard quadratic form

Rearrange the equation \(3v^2 + 2v = 1\) into the standard form of a quadratic equation: \(3v^2 + 2v - 1 = 0\).

Step 4: Use the quadratic formula

The standard quadratic equation is \(ax^2 + bx + c = 0\), where \(a = 3\), \(b = 2\), and \(c = -1\). The quadratic formula \(v = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\) solves this.

Step 5: Calculate the discriminant

Calculate the discriminant \(b^2 - 4ac\). Here, \(b^2 - 4ac = 2^2 - 4(3)(-1) = 4 + 12 = 16\).

Step 6: Solve for v using the quadratic formula

Substitute the values into the quadratic formula: \(v = \frac{{-2 \pm \sqrt{16}}}{6}\). This leads to \(v = \frac{{-2 \pm 4}}{6}\).

Step 7: Find the roots for v

Compute the values for \(v\). These are \(v_1 = \frac{2}{6} = \frac{1}{3}\) and \(v_2 = \frac{-6}{6} = -1\).

Step 8: Back-substitute to find x

Since \(v = e^x\), solve \(e^x = \frac{1}{3}\) and \(e^x = -1\). The second equation \(e^x = -1\) has no real solutions since the exponential function is always positive.

Step 9: Solve for x

For \(e^x = \frac{1}{3}\), take the natural logarithm of both sides: \(x = \ln(\frac{1}{3})\).

Step 10: Simplify the solution

Thus, the solution for \(x\) is \(x = -\ln(3)\), as \(\ln(\frac{1}{3}) = -\ln(3)\).

Key Concepts

Exponential Functions

Exponential functions are a crucial component in mathematics, playing an essential role in modeling growth and decay processes. In an exponential equation, the variable is located in the exponent. For example, in the equation \( e^{2x} \), \( e \) is the base of the natural logarithm (approximately 2.718), and \( 2x \) is the exponent. Exponential functions have the form \( y = a \cdot b^x \). In this form:\

\
• \( a \) is a constant coefficient,\
• \( b \) is the base of the exponent,\
• \( x \) is the variable exponent.\

\Such equations often require transformations, making them easier to solve by using substitutions. For example, substituting \( v = e^x \) can transform the exponential equation into a polynomial or quadratic form, as illustrated in the initial exercise problem.

Quadratic Formula

The quadratic formula is a solution for quadratic equations of the form \( ax^2 + bx + c = 0 \). It is expressed as:\\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] \This formula provides the roots (solutions) of the quadratic equation by substituting the coefficients \( a \), \( b \), and \( c \). Quadratic equations are fundamental in algebra due to their pervasive presence in scientific calculations and real-world scenarios.

Using this formula requires calculating the discriminant, \( b^2 - 4ac \), which will help determine the nature of the roots – whether real or imaginary. Once the discriminant is determined, the quadratic formula gives both potential roots of the equation, which can then be back-substituted to solve original problems like the one given.

Natural Logarithm

The natural logarithm, denoted as \ln(x)\, is a function that uses the base \( e \) (approximately 2.718). It is the inverse of the exponential function \( e^x \). This means if \( y = e^x \), then \( x = \ln(y) \). Natural logarithms are particularly useful for solving equations where the variable is in the exponent. For example, to solve \( e^x = \frac{1}{3} \), you take the natural logarithm of both sides to simplify it to \( x = \ln(\frac{1}{3}) \).

The property that \( \ln(ab) = \ln(a) + \ln(b) \) and \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \) can simplify complex logarithmic expressions. This is used in solving back-substitutions for the variable \( x \). When solving exponential equations, taking the natural logarithm can often reveal the solution in a much simpler form.

Discriminant

The discriminant is a part of the quadratic formula, critical for understanding the nature of the roots of a quadratic equation. It is given by the term \( b^2 - 4ac \), where \( a \), \( b \), and \( c \) are the coefficients of the terms in the quadratic equation \( ax^2 + bx + c = 0 \).

The discriminant tells us:\

\
• If it is positive, there are two distinct real solutions,\
• If it is zero, there is exactly one real solution (repeated),\
• If it is negative, there are no real solutions (solutions are complex or imaginary).\

\In our step-by-step solution, calculating \( 16 \) for the discriminant indicates that there are two real roots for the transformed equation. This information helps determine the feasibility and nature of solutions for \( v \), and consequently for the original variable \( x \) in the exponential equation.