Problem 127 - Elementary and Intermediate Algebra

Question

Solve each equation. $$\log _{3} x+\log _{3}(x+2)=2$$

Step-by-Step Solution

Answer

The solution is \(x = \sqrt{10} - 1\).

Step 1: Use Logarithm Properties to Combine Logs

Use the property of logarithms, \(\log_b(m) + \log_b(n) = \log_b(m \cdot n)\), to combine the logarithms on the left side of the equation. This gives us \(\log_3(x(x+2)) = 2\).

Step 2: Convert the Logarithmic Equation to an Exponential Equation

Rewrite the logarithmic equation as an exponential equation using the property that \(\log_b(A) = C \) is equivalent to \(b^C = A\). Therefore, \(3^2 = x(x+2)\).

Step 3: Expand the Quadratic Expression

Simplify the equation from Step 2: \(9 = x^2 + 2x\) by expanding the right side.

Step 4: Rearrange into a Quadratic Equation

Transform the equation into standard quadratic form: \(x^2 + 2x - 9 = 0\).

Step 5: Solve the Quadratic Equation

Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = -9\), to solve for \(x\).

Step 6: Calculate the Discriminant

First, calculate the discriminant: \(b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-9) = 4 + 36 = 40\).

Step 7: Find the Roots Using the Quadratic Formula

Plug the discriminant into the quadratic formula to find \(x\): \(x = \frac{-2 \pm \sqrt{40}}{2} = \frac{-2 \pm 2\sqrt{10}}{2} = -1 \pm \sqrt{10}\). The potential solutions are \(x = \sqrt{10} - 1\) and \(x = -\sqrt{10} - 1\).

Step 8: Verify the Valid Solution(s)

Only non-negative values of \(x\) make sense since logs require positive arguments. Thus, verify that \(x = \sqrt{10} - 1 > 0\). \(x = -\sqrt{10} - 1\) is negative and invalid.

Key Concepts

Properties of Logarithms

Logarithms have incredible properties that make them unique and useful in simplifying complex expressions. One fundamental property is the **Product Rule**. This rule is used when you have the logarithm of a product. It states:

• \( \log_b(m \, n) = \log_b(m) + \log_b(n) \)

This means you can combine two logarithms into one by multiplying the arguments inside the logs, provided they have the same base, as seen in the exercise: \( \log_3(x) + \log_3(x+2) = \log_3(x(x+2)) \). This simplifies complex expressions and is crucial when solving logarithmic equations.
Another important property is the **Power Rule**:

• \( \log_b(m^n) = n \times \log_b(m) \)

This rule allows you to take the exponent out of the logarithm, which can be handy in numerous calculations. Mastering these properties will give you the tools to tackle more complicated logarithmic challenges.

Exponential Equations

Exponential equations involve expressions where the variable is in the exponent. They often arise from logarithmic equations during their transformation. In our original exercise, we used this concept when we transformed \( \log_3(x(x+2)) = 2 \) into an exponential form.
By understanding that a logarithmic equation \( \log_b(A) = C \) is equivalent to an exponential form \( b^C = A \), we converted it to \( 3^2 = x(x+2) \), which simplified further problem-solving steps.
The solution process usually involves:

• Rewriting the logarithmic equation using its exponential form.
• Solving for the variable using traditional techniques like factoring or applying the quadratic formula.

Exponential equations are frequently encountered in real-world applications, from population growth calculations to understanding compound interest.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, equations of the form \( ax^2 + bx + c = 0 \). In our exercise, once the equation was rearranged to \( x^2 + 2x - 9 = 0 \), the quadratic formula was used. The formula is:

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

This method allows you to find the roots of any quadratic equation, even when factoring is difficult or impossible.
The formula involves:

• Substituting the coefficients \(a\), \(b\), and \(c\), which are derived from the equation in standard form.
• Calculating the discriminant (\(b^2 - 4ac\)), which determines the nature of the roots.

Remember, using the quadratic formula requires attention to detail in solving for both potential roots.

Discriminant in Quadratic Equations

The discriminant is a component of the quadratic formula that reveals critical information about the nature of the equation's roots. Calculated as \( b^2 - 4ac \), it determines the kind and number of solutions for a quadratic equation:

• If the discriminant is positive, the equation has two different real roots.
• If it's zero, the equation has exactly one real root, meaning the roots are equal.
• If it's negative, the equation has no real roots, only complex or imaginary roots.

In our original exercise with \( b = 2, a = 1, c = -9 \), the discriminant was calculated as 40, which is positive.
Thus, the equation has two distinct real roots, indicating the possible solutions \( x = \sqrt{10} - 1 \) and \( x = -\sqrt{10} - 1 \). This understanding is vital while solving quadratic equations, ensuring you analyze the correct number and types of solutions.