Problem 24 - Algebra and Trigonometry with Analytic Geometry

Question

Solve the equation. $$ \ln x+\ln (x+6)=\frac{1}{2} \ln 9 $$

Step-by-Step Solution

Answer

The solution is \( x = -3 + 2\sqrt{3} \).

Step 1: Use Logarithm Properties

The equation given is \( \ln x + \ln (x+6) = \frac{1}{2} \ln 9 \). The property of logarithms \( \ln a + \ln b = \ln (ab) \) can be used here. Therefore, the left side of the equation becomes \( \ln (x(x + 6)) \). So, the equation becomes \( \ln (x(x + 6)) = \frac{1}{2} \ln 9 \).

Step 2: Eliminate Logarithms

To simplify, remove the logarithms by exponentiating both sides: \( e^{\ln (x(x + 6))} = e^{\frac{1}{2} \ln 9} \). This simplifies to \( x(x + 6) = 9^{1/2} \). Since \( 9^{1/2} = 3 \), the equation is now \( x(x + 6) = 3 \).

Step 3: Expand and Rearrange

Expand the equation: \( x^2 + 6x = 3 \). Rearrange it to form a standard quadratic equation: \( x^2 + 6x - 3 = 0 \).

Step 4: Solve the Quadratic Equation Using Quadratic Formula

To solve the quadratic equation \( x^2 + 6x - 3 = 0 \), use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 6, \text{ and } c = -3 \). Substitute these into the equation: \( x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1} \).

Step 5: Calculate the Discriminant

The discriminant \( b^2 - 4ac \) is \( 6^2 - 4 \times 1 \times (-3) = 36 + 12 = 48 \). Thus, \( x = \frac{-6 \pm \sqrt{48}}{2} \).

Step 6: Simplify the Result

Since \( \sqrt{48} = 4\sqrt{3} \), substitute this back to obtain \( x = \frac{-6 \pm 4\sqrt{3}}{2} \). Simplifying gives two solutions: \( x = -3 + 2\sqrt{3} \) and \( x = -3 - 2\sqrt{3} \).

Step 7: Confirm Valid Solutions

Since \( x \) must be greater than 0 (as it was in the logarithm’s argument), \( x = -3 - 2\sqrt{3} \) is not valid because it is negative. Hence, the only valid solution is \( x = -3 + 2\sqrt{3} \).

Key Concepts

Logarithm Properties

Logarithms have several fundamental properties that make simplifying and solving equations easier. One essential property is the product rule:

• The property states that \( \ln a + \ln b = \ln (ab) \).

This property helps us combine logarithmic expressions, making the equation handling more straightforward. For instance, in our exercise, \( \ln x + \ln (x+6) \) is transformed into \( \ln (x(x + 6)) \). The equation becomes easier to manage, paving the way for the next steps in the solving process. Another helpful property is the power rule of logarithms which isn’t directly used here but is useful to know:

• \( \ln a^b = b \cdot \ln a \)

Knowing these properties is vital for efficiently manipulating logarithmic functions in various mathematical problems.

Quadratic Formula

When solving quadratic equations, the quadratic formula is an invaluable tool. For any equation of the form \( ax^2 + bx + c = 0 \), the solution can be found using:

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

This formula is powerful because it offers a direct way to determine the roots of any quadratic equation, whether the equation is easily factorable or not. In our exercise, after rearranging the equation \( x(x + 6) = 3 \) to \( x^2 + 6x - 3 = 0 \), we applied the quadratic formula to find the value of \( x \). It's widely used because it works in all cases where the quadratic is non-degenerate (\( a eq 0 \)), thus providing certainty in results. The formula involves computing the discriminant, which we will discuss next.

Discriminant Calculation

The discriminant is an important part of the quadratic formula, given by \( b^2 - 4ac \). It determines the nature of the roots of a quadratic equation. Here is what the discriminant tells us about the solutions:

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, the equation has exactly one real root (a repeated root).
• If it is negative, the equation has no real roots (the roots are complex).

In our example, we calculated the discriminant as \( 6^2 - 4 \times 1 \times (-3) = 48 \). Since 48 is positive, the quadratic equation has two distinct real roots. Calculating the discriminant is a crucial step because it provides insight into the nature and relationship of the equation's solutions.

Exponential Functions

Exponential functions play a key role when dealing with logarithms, especially when you need to eliminate them to solve equations. In our exercise, we used the property that \( e^{\ln a} = a \) to eliminate the logarithms. By exponentiating both sides of the equation \( \ln (x(x + 6)) = \frac{1}{2} \ln 9 \), we converted the logarithmic equation into a polynomial form. Exponential functions of the form \( y = e^x \) are vital in reversing the effect of logarithms in an equation. Remember:

• The base of natural logarithms is \( e \), approximately equal to 2.718.
• Exponential functions grow quickly and are used extensively in mathematical modeling.

Understanding how to handle exponential functions and their interplay with logarithms is essential for solving equations like the one we examined, which involves reducing the complexity of expressions to simpler algebraic forms.