Problem 65 - Precalculus

Question

Solve the logarithmic equations. Round your answers to three decimal places. $$\ln (x)+\ln (x-2)=4$$

Step-by-Step Solution

Answer

The solution is \( x = 8.456 \).

Step 1: Apply Logarithmic Properties

Use the property of logarithms that states \( \ln(a) + \ln(b) = \ln(ab) \). This means we can combine the two logarithmic expressions on the left-hand side: \( \ln(x) + \ln(x-2) \) becomes \( \ln(x(x-2)) \) or \( \ln(x^2 - 2x) \). Thus, the equation becomes: \( \ln(x^2 - 2x) = 4 \).

Step 2: Exponentiate to Remove the Logarithm

To solve the equation \( \ln(x^2 - 2x) = 4 \), we exponentiate both sides to eliminate the natural logarithm. This gives us: \( e^{\ln(x^2 - 2x)} = e^4 \). Since \( e^{\ln(a)} = a \), it follows that \( x^2 - 2x = e^4 \).

Step 3: Solve the Quadratic Equation

Now we have the quadratic equation: \( x^2 - 2x = e^4 \). Rearrange it to standard form: \( x^2 - 2x - e^4 = 0 \). Use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -e^4 \). Thus, \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-e^4)}}{2 \cdot 1} \).

Step 4: Calculate the Discriminant

Compute the discriminant: \( (-2)^2 - 4 \cdot 1 \cdot (-e^4) = 4 + 4e^4 \). Substitute \( e^4 \approx 54.598 \) to find: \( 4 + 4 \times 54.598 = 222.392 \).

Step 5: Use the Quadratic Formula

Substitute the discriminant back into the quadratic formula: \( x = \frac{2 \pm \sqrt{222.392}}{2} \). Compute \( \sqrt{222.392} \approx 14.912 \). Therefore, the potential solutions are \( x = \frac{2 + 14.912}{2} \) or \( x = \frac{2 - 14.912}{2} \).

Step 6: Compute Values and Verify

Compute the solutions: \( x = \frac{16.912}{2} = 8.456 \) and \( x = \frac{-12.912}{2} = -6.456 \). Since \( x \) must be greater than 2 for \( \ln(x-2) \) to be defined, only \( x = 8.456 \) is valid.

Key Concepts

Properties of Logarithms

Logarithms are a powerful tool in mathematics, especially when simplifying expressions or solving equations. One property of logarithms commonly used is the sum property. This property states that the logarithm of a product is equal to the sum of the logarithms:

• \( \ln(a) + \ln(b) = \ln(ab) \)

This property allows you to combine multiple logarithmic terms into a single logarithm, which simplifies the solving process. Another important property is the change of base formula, though it’s not used in this particular problem. By understanding these properties, you can manipulate and solve logarithmic equations with greater ease. Remember: mastering these properties can also aid in solving exponential equations by logarithmic conversions.

Quadratic Formula

The quadratic formula is essential for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). The formula itself is:

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

You can find the roots of any quadratic equation using this formula. The expression under the square root sign, \( b^2 - 4ac \), is known as the discriminant.

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If it's zero, there’s exactly one real root (a repeated root).
• A negative discriminant means the equation has two complex roots.

In our exercise, the discriminant was indeed positive, allowing us to find two potential real roots. However, it’s always crucial to check each for any constraints set by the context, such as domain restrictions in logarithmic equations.

Natural Logarithm

The natural logarithm, denoted as \( \ln(x) \), is a logarithm with base \( e \), where \( e \approx 2.71828 \). It's widely used in calculus and other mathematical fields due to its properties and natural behavior in growth processes.
Natural logarithms help in simplifying calculations that involve exponential growth laws or radioactive decay.
The logarithm property \( \ln(e^y) = y \) because \( e^y \) and \( y \) are inverse functions — they "undo" each other.In the problem at hand, exponentiating both sides (\( e^{\ln(a)} = a \)) was a crucial step in eliminating the natural logarithm and thus simplifying the equation to solve.Understanding this functional relationship is pivotal whenever shifts between logarithmic and exponential expressions are needed.

Exponentiation

Exponentiation refers to raising a number to the power of an exponent. For example, in \( e^x \), \( e \) is the base, and \( x \) is the exponent.
In mathematics, exponentiation is the inverse of taking logarithms — they are two sides of a conceptual coin. This feature is used when you want to "remove" a logarithm from an equation, as shown in solving \( \ln(x^2 - 2x) = 4 \) by exponentiating both sides to isolate the polynomial.Here's a quick refresher on basic exponent rules that help in simplification:

• \( a^{m+n} = a^m \times a^n \)
• \( (a^m)^n = a^{mn} \)
• \( a^{-n} = \frac{1}{a^n} \)

In advanced problems, knowing these rules and how they interact with logarithmic properties expedites solving challenging expressions.