Solving Ln(x+5) + Ln(x-1) = 2: A Step-by-Step Guide

by Alex Johnson 52 views

Introduction

In this article, we will walk through the process of solving the logarithmic equation ln(x+5) + ln(x-1) = 2. Logarithmic equations often appear complex, but by applying the properties of logarithms and some algebraic manipulation, we can find the solution. This step-by-step guide aims to provide a clear and comprehensive understanding of how to tackle such equations. Understanding logarithmic equations is crucial, especially if you're involved in fields like physics, engineering, or computer science, where these concepts are frequently applied. Therefore, let's dive in and unravel the solution to this equation together.

Step 1: Combine the Logarithms

The first step in solving the equation ln(x+5) + ln(x-1) = 2 is to combine the two logarithms on the left side into a single logarithm. We can use the property that the sum of two logarithms is the logarithm of the product of their arguments. Mathematically, this property is expressed as:

ln(a) + ln(b) = ln(ab)

Applying this property to our equation, we get:

ln((x+5)(x-1)) = 2

This simplifies the equation by reducing two logarithmic terms into one. Expanding the product inside the logarithm, we have:

ln(x² + 4x - 5) = 2

Combining logarithms simplifies the equation and makes it easier to work with in subsequent steps. This is a fundamental technique when solving logarithmic equations, so make sure to remember it. Always look for opportunities to combine or separate logarithms based on the properties available to you.

Step 2: Convert to Exponential Form

Now that we have a single logarithm, the next step is to convert the logarithmic equation into its equivalent exponential form. Recall that the natural logarithm, denoted as ln, has a base of e, where e is approximately 2.71828. The general relationship between logarithms and exponentials is:

ln(a) = b <=> e^b = a

Applying this to our equation ln(x² + 4x - 5) = 2, we get:

e² = x² + 4x - 5

This conversion eliminates the logarithm, transforming the equation into a more manageable algebraic form. The exponential form allows us to directly solve for x using algebraic techniques. The number e is a mathematical constant that appears in many contexts, especially in calculus and exponential growth/decay problems. Recognizing and utilizing this constant is essential for solving equations involving natural logarithms.

Step 3: Rearrange into a Quadratic Equation

After converting the equation to exponential form, we now have:

e² = x² + 4x - 5

To solve for x, we need to rearrange this equation into a standard quadratic form, which is:

ax² + bx + c = 0

Subtracting e² from both sides of our equation, we get:

x² + 4x - 5 - e² = 0

Thus, the quadratic equation we need to solve is:

x² + 4x - (5 + e²) = 0

Rearranging into quadratic form is crucial because it allows us to apply standard methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. The coefficients a, b, and c in the quadratic equation are essential for these methods. In our case, a = 1, b = 4, and c = -(5 + e²).

Step 4: Apply the Quadratic Formula

Now that we have the quadratic equation x² + 4x - (5 + e²) = 0, we can use the quadratic formula to find the values of x. The quadratic formula is given by:

x = (-b ± √(b² - 4ac)) / (2a)

In our equation, a = 1, b = 4, and c = -(5 + e²). Plugging these values into the quadratic formula, we get:

x = (-4 ± √(4² - 4(1)(-(5 + e²)))) / (2(1))

Simplifying the expression under the square root:

x = (-4 ± √(16 + 4(5 + e²))) / 2 x = (-4 ± √(16 + 20 + 4e²)) / 2 x = (-4 ± √(36 + 4e²)) / 2 x = (-4 ± √(4(9 + e²))) / 2 x = (-4 ± 2√(9 + e²)) / 2 x = -2 ± √(9 + e²)

So we have two possible solutions for x:

x₁ = -2 + √(9 + e²) x₂ = -2 - √(9 + e²)

The quadratic formula is a powerful tool for solving any quadratic equation and is particularly useful when factoring is not straightforward. It provides a direct method for finding the roots of the equation based on its coefficients. Remember to carefully substitute the values of a, b, and c to avoid errors in your calculation.

Step 5: Check for Extraneous Solutions

After finding the possible values of x, it’s crucial to check whether these solutions are valid by substituting them back into the original logarithmic equation ln(x+5) + ln(x-1) = 2. Logarithmic functions are only defined for positive arguments. Therefore, we must ensure that x+5 > 0 and x-1 > 0 for the solutions to be valid.

Let's analyze the two possible solutions:

  1. x₁ = -2 + √(9 + e²)

    Since e² is approximately 7.389, then √(9 + e²) ≈ √(9 + 7.389) ≈ √16.389 ≈ 4.048. So, x₁ ≈ -2 + 4.048 ≈ 2.048. Let's check if this value satisfies the conditions:

    • x₁ + 5 ≈ 2.048 + 5 ≈ 7.048 > 0
    • x₁ - 1 ≈ 2.048 - 1 ≈ 1.048 > 0 Thus, x₁ is a valid solution.

  2. x₂ = -2 - √(9 + e²)

    Using the same approximation, x₂ ≈ -2 - 4.048 ≈ -6.048. Let's check if this value satisfies the conditions:

    • x₂ + 5 ≈ -6.048 + 5 ≈ -1.048 < 0
    • x₂ - 1 ≈ -6.048 - 1 ≈ -7.048 < 0 Since x₂ + 5 and x₂ - 1 are both negative, x₂ is not a valid solution. It is an extraneous solution.

Therefore, the only valid solution is:

x = -2 + √(9 + e²)

Checking for extraneous solutions is a critical step in solving logarithmic equations. Always remember that the arguments of logarithms must be positive. By verifying each potential solution, we can avoid errors and ensure the accuracy of our results.

Conclusion

In summary, to solve the logarithmic equation ln(x+5) + ln(x-1) = 2, we followed these steps:

  1. Combined the logarithms using the property ln(a) + ln(b) = ln(ab).
  2. Converted the equation to exponential form using the relationship between logarithms and exponentials.
  3. Rearranged the equation into a standard quadratic form.
  4. Applied the quadratic formula to find the possible values of x.
  5. Checked for extraneous solutions by ensuring the arguments of the logarithms were positive.

By following these steps, we found that the only valid solution to the equation is:

x = -2 + √(9 + e²)

Understanding and applying these techniques will enable you to solve a wide variety of logarithmic equations. Remember to pay close attention to the properties of logarithms and to always check for extraneous solutions to ensure accuracy. Further your understanding by checking out resources like Khan Academy's section on logarithmic equations: Khan Academy Logarithmic Equations. This will help reinforce the concepts and skills needed to master these types of problems.