Solve Logarithmic Equations: A Step-by-Step Guide

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Hey there, math enthusiasts! Today, we're diving deep into the world of logarithmic equations. We're going to tackle a specific problem: solving for x in the equation $\log_2(x+3) + \log_2(x-3) = 4$. This might seem daunting at first, but don't worry, we'll break it down step by step, making it super easy to understand. So, grab your thinking caps, and let's get started!

Understanding Logarithms: The Basics

Before we jump into solving the equation, let's quickly recap what logarithms are all about. Logarithms are basically the inverse operation of exponentiation. Think of it this way: if $2^3 = 8$, then $\log_2(8) = 3$. The logarithm (base 2) of 8 is the exponent (3) to which you need to raise 2 to get 8.

In simpler terms, a logarithm answers the question: "What exponent do I need to put on this base to get this number?" Understanding this fundamental concept is key to conquering logarithmic equations. You see, the main concept behind logarithms is simply to undo exponentiation. It is because they're inverses that they get their amazing solving power. This is how we're able to solve exponential equations and, as we're seeing, logarithmic equations. Remember the properties of logs like the product rule, quotient rule, and power rule? They will be very helpful when solving these equations.

When working with logarithms, it's crucial to remember the domain. The argument of a logarithm (the thing inside the parentheses) must always be positive. This is because you can't raise a positive base to any power and get a negative number or zero. This little detail will be very important when we check our solutions later on.

So, with the basics under our belt, we're ready to tackle the equation at hand. We'll use the properties of logarithms to simplify the equation, solve for x, and then make sure our solution makes sense within the domain. It's like a mathematical adventure, and we're the explorers! Remember, practice makes perfect, so the more you work with logarithms, the more comfortable you'll become. Keep an eye out for those domain restrictions – they're the secret guardians of logarithmic solutions!

Step 1: Condensing the Logarithmic Expression

The first thing we need to do is simplify the equation. Notice that we have two logarithms added together on the left side. This is where the product rule of logarithms comes into play. The product rule states that: $\log_b(m) + \log_b(n) = \log_b(mn)$.

Guys, this rule is a lifesaver! It allows us to combine two logarithms with the same base into a single logarithm. In our case, the base is 2, so we can rewrite the equation as:

$\log_2((x+3)(x-3)) = 4$

See how we've combined the two logarithms into one? This makes the equation much easier to work with. Now, let's simplify the expression inside the logarithm. Remember the difference of squares? $(a+b)(a-b) = a^2 - b^2$. Applying this to our equation, we get:

$\log_2(x^2 - 9) = 4$

Awesome! We've successfully condensed the logarithmic expression into a single logarithm. This is a crucial step in solving logarithmic equations. By using the product rule, we've transformed a complex-looking equation into a simpler one. Now, we're one step closer to finding the value of x. The beauty of math lies in these clever transformations that make seemingly impossible problems solvable. So, keep these logarithmic properties in mind – they're your best friends in the world of logarithms!

Remember, the key is to recognize the patterns and apply the appropriate rules. In this case, spotting the sum of two logarithms immediately triggers the thought of the product rule. It's like having a secret weapon in your mathematical arsenal! With each step, we're peeling away the layers of the problem, revealing the solution underneath. So, let's move on to the next step and unleash our mathematical prowess!

Step 2: Converting to Exponential Form

Now that we have a single logarithm, we need to get rid of it to isolate x. This is where we use the relationship between logarithms and exponents. Remember that $\log_b(a) = c$ is equivalent to $b^c = a$. We're simply rewriting the equation in exponential form.

In our case, we have $\log_2(x^2 - 9) = 4$. Applying the conversion rule, we get:

$2^4 = x^2 - 9$

Boom! We've successfully converted the logarithmic equation into an exponential equation. This is a major breakthrough because we now have a familiar algebraic equation to solve. No more logarithms to worry about! This step highlights the power of understanding the inverse relationship between logarithms and exponents. It's like having a translator that can switch between two different mathematical languages.

Now, let's simplify the equation further. We know that $2^4 = 16$, so we have:

$16 = x^2 - 9$

We're getting closer and closer to finding the value of x. Each step we take simplifies the equation, making it more manageable. This is the essence of problem-solving – breaking down a complex problem into smaller, easier-to-solve steps. Now that we have a simple quadratic equation, we can use our algebraic skills to find the solutions for x. So, let's move on to the next step and conquer this equation!

Remember, the key to success in mathematics is to connect the dots between different concepts. In this step, we used the fundamental relationship between logarithms and exponents to transform the equation. It's like building a bridge between two islands, allowing us to cross over and reach our destination. So, keep these connections in mind, and you'll become a master problem-solver!

Step 3: Solving the Quadratic Equation

We've transformed our logarithmic equation into a simple quadratic equation: $16 = x^2 - 9$. Now it's time to solve for x. The first thing we need to do is get all the terms on one side of the equation, setting it equal to zero. Let's add 9 to both sides:

$x^2 - 9 + 9 = 16 + 9$

This simplifies to:

$x^2 = 25$

Alright! We're almost there. Now, to isolate x, we need to take the square root of both sides. Remember, when we take the square root, we need to consider both positive and negative solutions:

$\sqrt{x^2} = \pm \sqrt{25}$

This gives us:

$x = \pm 5$

So, we have two potential solutions: $x = 5$ and $x = -5$. But hold on, we're not done yet! We need to check these solutions to make sure they're valid.

Solving quadratic equations is a fundamental skill in algebra, and it's crucial for tackling many mathematical problems. In this case, we used the square root property to find the solutions. Remember to always consider both positive and negative roots when taking the square root. It's a common mistake to forget the negative root, which can lead to an incomplete solution. So, keep this in mind as you solve quadratic equations in the future!

We've come a long way in solving this problem, from condensing the logarithmic expression to converting it to exponential form and finally solving the quadratic equation. Each step has brought us closer to the solution, and now we have two potential answers. But before we celebrate, we need to make sure these solutions are valid within the context of the original equation. So, let's move on to the final step: checking our solutions!

Step 4: Checking for Extraneous Solutions

Remember how we talked about the domain of logarithms? This is where it comes into play. The argument of a logarithm must always be positive. So, we need to check if our solutions, $x = 5$ and $x = -5$, satisfy this condition in the original equation: $\log_2(x+3) + \log_2(x-3) = 4$.

Let's start with $x = 5$. Plugging it into the equation, we get:

$\log_2(5+3) + \log_2(5-3) = \log_2(8) + \log_2(2)$

Both arguments, 8 and 2, are positive, so this solution looks promising. Now, let's check if it satisfies the equation:

$\log_2(8) + \log_2(2) = 3 + 1 = 4$

Great! $x = 5$ is a valid solution.

Now, let's check $x = -5$. Plugging it into the equation, we get:

$\log_2(-5+3) + \log_2(-5-3) = \log_2(-2) + \log_2(-8)$

Uh oh! We have negative arguments inside the logarithms. This means $x = -5$ is an extraneous solution. It doesn't satisfy the domain of the logarithms, so we must discard it.

This is a crucial step, guys! Always check your solutions in logarithmic equations. Extraneous solutions can sneak in due to the nature of logarithms, so it's essential to verify that your answers are valid.

Checking for extraneous solutions is like being a detective, making sure all the pieces of the puzzle fit together. We found two potential solutions, but only one of them held up under scrutiny. This step highlights the importance of rigor in mathematics – we can't just stop at finding potential solutions; we need to verify them to ensure accuracy.

So, after carefully checking our solutions, we've determined that only $x = 5$ is a valid solution. We've successfully navigated the world of logarithmic equations and emerged victorious! Now, let's write down our final answer.

Final Answer:

The solution to the equation $\log_2(x+3) + \log_2(x-3) = 4$ is:

A. The solution(s) is/are $x = 5$.

Woohoo! We did it! We successfully solved the logarithmic equation and found the correct solution. Remember, the key to solving logarithmic equations is to understand the properties of logarithms, convert the equation to exponential form, solve the resulting algebraic equation, and most importantly, check for extraneous solutions.

By following these steps, you can conquer any logarithmic equation that comes your way. Keep practicing, and you'll become a logarithm-solving pro in no time!

So, there you have it, folks! A comprehensive guide to solving logarithmic equations, with a specific example tackled step by step. Remember, math is like a puzzle – each step is a piece that fits together to reveal the final solution. Keep exploring, keep learning, and keep solving! And most importantly, have fun with it! This journey through logarithms has shown us the power of mathematical thinking and the satisfaction of finding the right answer. So, go forth and conquer more mathematical challenges!