Logarithmic Equation Solution: Extraneous Roots Explained

Introduction

Hey guys! Today, we're diving deep into the fascinating world of logarithmic equations. We'll be tackling a problem that Janet solved, and it's a classic example that highlights the importance of understanding the nuances of logarithmic functions. Janet's equation is $\log (x-3)+\log x=1$, and she arrived at two potential solutions: $x=5$ and $x=-2$. But here's the catch: not all solutions we find in the process of solving equations are actually valid. These sneaky imposters are called extraneous solutions, and they're a common pitfall when dealing with logarithms. So, let's break down Janet's work, explore why extraneous solutions pop up, and learn how to identify and avoid them. This journey will not only help you ace your math tests but also give you a solid understanding of the underlying principles. We'll make sure to cover every step in detail, so even if you're just starting out with logarithms, you'll be able to follow along. So, buckle up, grab your thinking caps, and let's get started!

Janet's Equation and Potential Solutions

Let's begin by revisiting the equation Janet tackled: $\log (x-3)+\log x=1$. This equation combines two logarithmic terms, and our goal is to isolate x. To do this effectively, we need to remember the fundamental properties of logarithms. One crucial property is the product rule, which states that the sum of logarithms is equal to the logarithm of the product. In mathematical terms, $\log_b{m} + \log_b{n} = \log_b{mn}$. Applying this rule to Janet's equation, we can combine the two logarithmic terms into a single logarithm:

$$\log[(x-3)x] = 1$$

Now, we have a simpler equation to work with. To get rid of the logarithm, we need to understand the relationship between logarithms and exponents. Remember that the logarithmic equation $\log_b{a} = c$ is equivalent to the exponential equation $b^c = a$. In our case, we're dealing with a common logarithm (base 10), so we can rewrite the equation as:

$$10^1 = (x-3)x$$

This simplifies to:

$$10 = x^2 - 3x$$

Rearranging the terms, we get a quadratic equation:

$$x^2 - 3x - 10 = 0$$

Now, we need to solve this quadratic equation. There are a couple of ways to do this: factoring or using the quadratic formula. In this case, the equation is easily factorable. We're looking for two numbers that multiply to -10 and add up to -3. These numbers are -5 and 2. So, we can factor the quadratic equation as:

$$(x - 5)(x + 2) = 0$$

This gives us two potential solutions:

$$x = 5 \quad \text{or} \quad x = -2$$

These are the solutions Janet found. But as we mentioned earlier, we're not done yet. We need to check if these solutions are valid in the original logarithmic equation.

The Crucial Step: Checking for Extraneous Solutions

This is where the real magic happens! We've found two potential solutions, but we need to make sure they actually work in the original equation. Remember, logarithms are only defined for positive arguments. This means that the expressions inside the logarithms, (x-3) and x, must be greater than zero. This is a critical concept when dealing with logarithmic equations.

Let's start by testing $x = 5$. Plugging this value into the original equation, we get:

$$\log(5 - 3) + \log(5) = 1$$

$$\log(2) + \log(5) = 1$$

Using the product rule again, we have:

$$\log(2 \cdot 5) = 1$$

$$\log(10) = 1$$

Since $\log_{10}(10) = 1$, the equation holds true. So, $x = 5$ is a valid solution. Great! We've got one confirmed solution.

Now, let's test $x = -2$. Plugging this value into the original equation, we get:

$$\log(-2 - 3) + \log(-2) = 1$$

$$\log(-5) + \log(-2) = 1$$

Here's the problem: we have logarithms of negative numbers. As we discussed earlier, logarithms are not defined for negative arguments. This means that $x = -2$ is not a valid solution. It's an extraneous solution! It emerged during the algebraic process of solving the equation, but it doesn't satisfy the original equation's domain restrictions.

This is why checking your solutions is absolutely crucial when dealing with logarithmic equations. Failing to do so can lead to incorrect answers and a misunderstanding of the problem.

Why Extraneous Solutions Occur

Understanding why extraneous solutions occur is just as important as knowing how to identify them. Extraneous solutions often arise because we manipulate equations in ways that expand the domain of possible solutions. In the case of logarithmic equations, we use properties like the product rule to combine logarithmic terms. While these properties are valid, they can sometimes mask the original domain restrictions.

For example, when we combined $\log(x-3) + \log(x)$ into $\log[(x-3)x]$, we essentially changed the domain. In the original equation, both (x-3) and x had to be positive. However, in the combined form, only the product (x-3)x needs to be positive. This allows for scenarios where one of the factors is negative, as long as the other factor is also negative, resulting in a positive product. This is precisely what happened with $x = -2$. It made the product (x-3)x positive, but it made the individual arguments of the logarithms negative, violating the original domain restrictions.

Think of it like this: imagine you're trying to find the entrance to a secret garden. The original logarithmic equation is like a map with very specific instructions, ensuring you only reach the true entrance. When we manipulate the equation, we're essentially redrawing the map. Sometimes, our new map might lead us to a false entrance – an extraneous solution – that looks like the real deal but ultimately doesn't work.

Therefore, always remember that the act of solving an equation can sometimes introduce solutions that weren't there in the beginning. This is why the checking step is so vital in the process of solving logarithmic (and other types of) equations.

Janet's Conclusion and the Importance of Domain

So, to wrap up Janet's problem, we can confidently state that of Janet's two solutions, only x = 5 is a valid solution because x = -2 results in taking the logarithm of a negative number, which is undefined.

This exercise underscores the fundamental importance of considering the domain of logarithmic functions. The domain is the set of all possible input values (in this case, x) for which the function is defined. For logarithmic functions, the domain is restricted to positive numbers. This restriction is a direct consequence of the definition of logarithms as the inverse of exponential functions. Exponential functions always produce positive outputs, so their inverse, the logarithm, can only accept positive inputs.

Understanding the domain of a function is not just a technicality; it's a core concept that affects how we solve equations and interpret results. By keeping the domain in mind, we can avoid pitfalls like extraneous solutions and ensure that our answers are meaningful and correct.

Tips for Mastering Logarithmic Equations

Alright, guys, you've made it this far, and you're well on your way to becoming logarithmic equation masters! To solidify your understanding, here are some key tips to keep in mind:

1. Know Your Logarithmic Properties Inside and Out: We've already talked about the product rule, but there's also the quotient rule, the power rule, and the change of base formula. These are your tools of the trade, so make sure you're comfortable using them.

2. Always Check for Extraneous Solutions: I can't stress this enough! Make it a habit to plug your potential solutions back into the original equation and verify that they work. It's a small step that can save you from big mistakes.

3. Pay Attention to the Domain: Remember that the arguments of logarithms must be positive. Keep this in mind throughout the solving process and use it to eliminate potential solutions.

4. Practice, Practice, Practice: The more you work with logarithmic equations, the more comfortable you'll become. Start with simpler problems and gradually work your way up to more complex ones.

5. Visualize Logarithmic Functions: Understanding the graph of a logarithmic function can give you valuable insights into its behavior. Notice how the function is only defined for positive x-values and how it approaches negative infinity as x approaches zero.

6. Rewrite Logarithmic Equations in Exponential Form: This is a useful method for solving a lot of equations

Conclusion

So, there you have it! We've explored Janet's equation, delved into the world of extraneous solutions, and reinforced the importance of understanding the domain of logarithmic functions. Logarithmic equations can seem daunting at first, but with a solid grasp of the fundamental principles and a healthy dose of practice, you can conquer them with confidence. Remember to always check your solutions, keep the domain in mind, and don't be afraid to ask for help when you need it. Keep practicing, keep exploring, and you'll be a logarithmic equation whiz in no time!

Math can be challenging, but with the right approach and a bit of perseverance, you can master it. So keep up the great work, and never stop learning!