Problem 93 - Precalculus

Question

Determine whether each statement is true or false. \(\log _{3}\left(x^{2}+x-6\right)=1\) has two solutions.

Step-by-Step Solution

Answer

True, the equation has two real solutions.

Step 1: Understand the Logarithmic Equation

The given equation is \(\log_{3}(x^2 + x - 6) = 1\). This means that the expression inside the logarithm \(x^2 + x - 6\) is equal to the base (3) raised to the power of 1.

Step 2: Rewrite the Equation in Exponential Form

Convert the logarithmic equation to its exponential form: \(x^2 + x - 6 = 3^1\), which simplifies to \(x^2 + x - 6 = 3\).

Step 3: Rearrange the Equation

Subtract 3 from both sides to form a quadratic equation: \(x^2 + x - 9 = 0\).

Step 4: Solve the Quadratic Equation

The quadratic equation \(x^2 + x - 9 = 0\) can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 1\), and \(c = -9\).

Step 5: Calculate the Discriminant

Calculate \(b^2 - 4ac\): \(1^2 - 4 \cdot 1 \cdot (-9) = 1 + 36 = 37\), which is positive, allowing for two real solutions.

Step 6: Apply the Quadratic Formula

Substitute the values into the quadratic formula: \(x = \frac{-1 \pm \sqrt{37}}{2}\). This results in two solutions: \(x_1 = \frac{-1 + \sqrt{37}}{2}\) and \(x_2 = \frac{-1 - \sqrt{37}}{2}\).

Key Concepts

Quadratic Equations

Quadratic equations are essential mathematical expressions that involve a variable raised to the second power. They typically come in the standard form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \) to ensure the equation remains quadratic rather than linear.
Understanding quadratic equations is crucial since they pop up in various mathematical problems.

• They can model real-world scenarios like projectile motion and areas.
• Solving them involves techniques such as factoring, completing the square, or using the quadratic formula.

In the example above, after rewriting the logarithmic equation into its exponential form, we rearrange it as a quadratic equation: \( x^2 + x - 9 = 0 \). Learning to identify and manipulate quadratic equations is foundational in algebra.

Exponential Form

Transforming a logarithmic equation into its exponential form is a key step in making it easier to solve. Logarithmic equations involve expressions like \( \log_{b}(y) = x \), indicating that the base \( b \) raised to the power of \( x \) equals \( y \).
This transformation is significant because it turns a potentially complex logarithmic equation into a more familiar polynomial or quadratic form, which can be solved through standard algebraic methods.

• In the original problem, converting \( \log_{3}(x^2 + x - 6) = 1 \) to exponential form leads to \( x^2 + x - 6 = 3 \).
• The resulting expression can then be simplified or rearranged into a manageable equation.

Mastering this conversion technique aids in tackling various algebra problems efficiently.

Discriminant

The discriminant is a vital component of the quadratic formula, playing a crucial role in determining the nature of the solutions to a quadratic equation. Denoted by \( b^2 - 4ac \), it helps us understand the characteristics of the solutions:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it is negative, the solutions are complex or imaginary numbers.

In the provided solution, the discriminant is calculated as \( 1^2 - 4 \cdot 1 \cdot (-9) = 37 \).
Because 37 is positive, the equation \( x^2 + x - 9 = 0 \) has two distinct real solutions, which is essential for confirming the statement's validity.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions to any quadratic equation, expressed as \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula uses coefficients \( a \), \( b \), and \( c \) from the standard quadratic equation \( ax^2 + bx + c = 0 \).
It's immensely valuable because it provides a straightforward way to obtain solutions without requiring factoring or completing the square.

• In our journey of solving \( x^2 + x - 9 = 0 \), we apply the quadratic formula using the coefficients: \( a = 1 \), \( b = 1 \), and \( c = -9 \).
• Substituting these values, we get the solutions \( x = \frac{-1 \pm \sqrt{37}}{2} \).

This results in two solutions: \( x_1 = \frac{-1 + \sqrt{37}}{2} \) and \( x_2 = \frac{-1 - \sqrt{37}}{2} \), affirming the reliability of the quadratic formula in solving any quadratic equation.