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

by Alex Johnson 47 views

Have you ever encountered a logarithmic equation that seemed daunting? Logarithmic equations might look intimidating at first, but with a clear understanding of logarithmic properties and algebraic techniques, they can be solved quite easily. In this guide, we'll break down the process of solving the equation ln(x-2) = ln(5x) step by step. Whether you're a student tackling homework or just brushing up on your math skills, this guide will provide you with a solid foundation. So, let’s dive into the fascinating world of logarithms and learn how to crack this equation!

Understanding Logarithms

Before we dive into solving the equation, it’s important to understand what logarithms are. At its core, a logarithm is the inverse operation to exponentiation. Think of it this way: if 2^3 = 8, then the logarithm base 2 of 8 is 3, written as log₂8 = 3. In simpler terms, the logarithm answers the question, “To what power must we raise the base to get this number?” Logarithms come in different forms, but the one we're focusing on today is the natural logarithm, denoted as "ln". The natural logarithm uses the base e, which is approximately 2.71828. So, when you see ln(x), it means "the power to which e must be raised to equal x.” Understanding this fundamental concept is the key to unlocking the secrets of logarithmic equations.

Key Properties of Logarithms

To effectively solve logarithmic equations, we need to be familiar with some key properties. These properties allow us to manipulate and simplify equations, making them easier to solve. Here are a few essential properties that will come in handy:

  1. Product Rule: ln(ab) = ln(a) + ln(b). This rule states that the logarithm of a product is the sum of the logarithms of the individual factors.
  2. Quotient Rule: ln(a/b) = ln(a) - ln(b). This rule says that the logarithm of a quotient is the difference between the logarithms of the numerator and the denominator.
  3. Power Rule: ln(a^b) = b * ln(a). This rule tells us that the logarithm of a number raised to a power is the product of the power and the logarithm of the number.
  4. One-to-One Property: If ln(a) = ln(b), then a = b. This property is particularly crucial for solving equations like the one we’re tackling today. It allows us to eliminate the logarithms and focus on the arguments (the expressions inside the logarithms).
  5. Inverse Property: e^(ln(x)) = x and ln(e^x) = x. These properties highlight the inverse relationship between the natural exponential function and the natural logarithm.

These properties will be our toolkit as we move forward, helping us transform complex logarithmic expressions into simpler forms. Make sure you have a good grasp of these rules before moving on, as they’re the backbone of logarithmic equation solving.

Solving the Equation ln(x-2) = ln(5x)

Now, let’s get to the heart of the matter and solve the equation ln(x-2) = ln(5x). This equation presents a classic scenario where we can apply the properties of logarithms to find the solution. Our primary goal is to isolate x, and we’ll do this by carefully applying logarithmic principles and algebraic techniques. Remember, solving logarithmic equations often involves a few key steps: simplifying the equation, eliminating the logarithms, solving the resulting algebraic equation, and, most importantly, checking for extraneous solutions. Extraneous solutions are values that satisfy the transformed equation but not the original one, so checking is a crucial step. Let's begin!

Step 1: Applying the One-to-One Property

The equation we have is ln(x-2) = ln(5x). Notice that we have a natural logarithm on both sides of the equation. This is where the One-to-One Property comes into play. As we discussed earlier, the One-to-One Property states that if ln(a) = ln(b), then a = b. This allows us to eliminate the natural logarithms from both sides and focus on the arguments inside the logarithms. Applying this property, we can rewrite our equation as:

x - 2 = 5x

This step is a game-changer! By using the One-to-One Property, we've transformed a logarithmic equation into a simple algebraic equation. This simplification is the key to unlocking the solution. Now, we can use basic algebraic techniques to isolate x.

Step 2: Solving the Algebraic Equation

Now that we have the equation x - 2 = 5x, we can proceed to solve for x. This involves isolating x on one side of the equation. Let’s start by subtracting x from both sides:

x - 2 - x = 5x - x

This simplifies to:

-2 = 4x

Next, we need to isolate x completely. To do this, we divide both sides of the equation by 4:

-2 / 4 = 4x / 4

This gives us:

x = -1/2

So, based on our algebraic manipulations, we have found a potential solution: x = -1/2. However, we're not done yet! It’s crucial to remember that we need to check this solution in the original logarithmic equation to make sure it’s valid.

Step 3: Checking for Extraneous Solutions

This is perhaps the most critical step in solving logarithmic equations. We must check our potential solution, x = -1/2, in the original equation, ln(x-2) = ln(5x), to ensure that it doesn't lead to any undefined logarithmic expressions. Remember, the argument of a logarithm must be positive. If we plug in x = -1/2 into the original equation, we get:

ln(-1/2 - 2) = ln(5 * -1/2)

Simplifying the arguments inside the logarithms:

ln(-2.5) = ln(-2.5)

Here’s the problem: The logarithm of a negative number is undefined. Logarithms are only defined for positive arguments. This means that x = -1/2 is an extraneous solution. It’s a solution that we obtained through our algebraic steps, but it does not satisfy the original logarithmic equation.

Step 4: Determining the Final Solution

Since we found that x = -1/2 is an extraneous solution, and it’s the only solution we found, this means that the original equation, ln(x-2) = ln(5x), has no solution. There is no value of x that can satisfy the equation without resulting in the logarithm of a negative number.

Therefore, the final answer is:

No Solution (∅)

This outcome highlights the importance of checking for extraneous solutions when solving logarithmic equations. It’s a step you should never skip!

Conclusion

Solving logarithmic equations can be a rewarding mathematical exercise. In this guide, we tackled the equation ln(x-2) = ln(5x) and walked through each step, from understanding the fundamental properties of logarithms to checking for extraneous solutions. We learned that by applying the One-to-One Property, we could transform the logarithmic equation into a simple algebraic equation. However, the crucial lesson here is that we must always check our solutions in the original equation to avoid extraneous results. In this case, our potential solution turned out to be extraneous, leading us to the conclusion that the equation has no solution.

Remember, mathematics is a journey of discovery, and every problem solved adds to your understanding. Keep practicing, and you’ll become more confident in your ability to tackle even the most complex logarithmic equations. And don't forget to explore further resources on logarithms and exponential functions for a deeper understanding.