Solving Logarithmic Equations: A Step-by-Step Guide

Introduction: Understanding Logarithms and Equations

Hey guys! Let's dive into the world of logarithms and equations. Today, we're tackling a math problem that might seem a bit tricky at first, but trust me, we'll break it down into easy-to-understand steps. Our mission? To solve the equation: $\log_3(x-1) + \log_3(x+1) = 1$. This is a classic example of a logarithmic equation, and by the end of this, you'll be a pro at solving these types of problems. So, what exactly are we dealing with? Logarithms are the inverse operations of exponentiation. They answer the question: "To what power must we raise a base to get a certain number?" In our equation, the base is 3. The equation asks us to find the value of $x$ that satisfies the given logarithmic expression. It's crucial to understand the properties of logarithms, particularly the product rule, which we'll be using extensively. This rule states that the sum of the logarithms of two numbers is equal to the logarithm of the product of those numbers. It's like a secret code that simplifies our equation and unlocks the solution. Keep in mind that we're looking for a solution that is accurate to three decimal places, so we need to be precise with our calculations. Let's embark on this journey together, and I promise, you'll feel confident in solving logarithmic equations by the time we're done! We'll be covering all the bases – the product rule, converting logarithms to exponential form, solving quadratic equations, and checking our answers to make sure they make sense in the original equation. So, get ready to flex those math muscles and let's get started. Ready? Let's go!

Step 1: Applying the Product Rule of Logarithms

Alright, first things first, let's simplify this equation. Our equation starts as $\log_3(x-1) + \log_3(x+1) = 1$. The first thing we should do is use the product rule of logarithms. Remember, this rule says that the sum of two logarithms with the same base is equal to the logarithm of the product of the arguments. So, we can rewrite our equation as: $\log_3((x-1)(x+1)) = 1$. See how we combined the two separate logs into one? It's a super handy trick that makes the equation much easier to handle. By applying the product rule, we've taken a big step toward solving the problem. What we've done is essentially collapsed the left side of the equation into a single logarithmic term, which will be important as we proceed. Think of it like this: the product rule is like a bridge that connects the two sides of the equation. We are connecting the two logarithms into a single expression, making the equation easier to manipulate and ultimately solve. Now, let's not get too carried away! We've made progress, but we still have a ways to go. But hey, at least it's starting to look a little less intimidating, right? We’re on our way to solving for $x$ and gaining mastery of the logarithmic equation. Hang in there! Trust me, these steps will become second nature before you know it.

Step 2: Converting the Logarithmic Equation to Exponential Form

Now that we've used the product rule, let's convert the logarithmic equation to exponential form. This is where things get a little more straightforward. The equation $\log_3((x-1)(x+1)) = 1$ translates to $3^1 = (x-1)(x+1)$. Remember, the base of the logarithm (which is 3 in our case) becomes the base of the exponent, and the result of the logarithm (which is 1) becomes the exponent. The argument of the logarithm, $(x-1)(x+1)$, becomes the result of the exponential expression. We are basically rewriting the problem in a different way, which will allow us to find the value of x. This step is crucial because it gets rid of the logarithm and turns the equation into something we can directly solve using more familiar algebraic techniques. Converting the logarithmic form to exponential form is like unlocking a door. We've now changed the equation from something we don't quite understand to something much easier to handle. Once you get the hang of this, solving logarithms will become much easier. Always remember: The base stays the base, the exponent is the result, and the argument is what the exponent equals. It's all about understanding this core relationship. Are you following? Great, let's keep moving. We are making fantastic progress, and pretty soon, we'll have our solution.

Step 3: Simplifying and Solving the Quadratic Equation

Okay, time to simplify that equation. From the previous step, we have $3^1 = (x-1)(x+1)$. First, calculate the left side of the equation, $3^1 = 3$, then, expand the right side of the equation, $(x-1)(x+1)$, this is a difference of squares. $(x-1)(x+1) = x^2 - 1$. Now, our equation looks like this: $3 = x^2 - 1$. Now we have a simple quadratic equation. Let's solve it. To get $x$ by itself, we will add 1 to both sides, we get $x^2 = 4$. Now, to isolate $x$, we take the square root of both sides, we have two possible values for x: $x = 2$ or $x = -2$. Whenever you're dealing with square roots, it's super important to remember that both positive and negative values are valid solutions. Good job, you're doing great! You have now solved for $x$!. But, hold on a sec, we're not quite done yet. Just because we've found two potential values for $x$ doesn't mean they both work in the original equation. We always have to check our answers when dealing with logarithms, because sometimes, we can get extraneous solutions (solutions that don't actually work). So, before we celebrate, we need to make sure these values are valid.

Step 4: Checking for Extraneous Solutions

Alright, time to see if our potential solutions, $x = 2$ and $x = -2$, actually work in our original equation: $\log_3(x-1) + \log_3(x+1) = 1$. The most important step is to determine if the values make sense within the context of the original problem. Because logarithms are only defined for positive arguments, this is where extraneous solutions can creep in. Let's start with $x = 2$. If we plug $x = 2$ into the equation, we get: $\log_3(2-1) + \log_3(2+1) = \log_3(1) + \log_3(3)$. Since the log of 1 with any base is 0, and the log of 3 with base 3 is 1, the equation becomes: $0 + 1 = 1$. This is correct! So, $x = 2$ is a valid solution. Great! Now, let's check $x = -2$. Plugging in $x = -2$, we get: $\log_3(-2-1) + \log_3(-2+1) = \log_3(-3) + \log_3(-1)$. Uh oh! Logarithms are not defined for negative numbers. Therefore, both $\log_3(-3)$ and $\log_3(-1)$ are undefined. This means that $x = -2$ is an extraneous solution and doesn't work in our equation. This happens sometimes with logarithms. So, it's super important to always check your answers. If we hadn't checked our solutions, we might have incorrectly said that -2 was a solution. Remember, logarithms only accept positive numbers as arguments. Always, always check your answers to make sure they are valid.

Step 5: The Final Answer and Conclusion

We've made it! After all the hard work, we've found the solution to our equation. We started with $\log_3(x-1) + \log_3(x+1) = 1$ and, after applying the product rule, converting to exponential form, solving, and checking for extraneous solutions, we found that $x = 2$ is the only valid solution. And it is accurate to three decimal places. Now, how cool is that? We took a tricky-looking logarithmic equation and broke it down step by step, transforming it into something we could easily solve. We learned about the product rule, converting between logarithmic and exponential forms, how to solve quadratic equations, and most importantly, how to check for extraneous solutions. Remember, the key to solving these types of problems is to understand the properties of logarithms and to take it one step at a time. Keep practicing, and you'll become a master of logarithmic equations in no time. So, the answer, accurate to three decimal places, is $x = 2.000$. Congratulations! You've successfully solved the equation. That was amazing, wasn't it? Now, go forth and conquer those logarithmic equations! You’ve earned it, guys!