Solving Logarithmic Equations Step-by-Step A Guide To Log(x^2-15)=log(2x)

Hey everyone! Today, we're diving deep into the fascinating world of logarithmic equations. Logarithmic equations might seem daunting at first, but trust me, with a systematic approach, you can conquer them like a math ninja! We're going to break down the equation $\log (x^2-15)=\log (2 x)$ step-by-step, so you'll not only understand how to solve it but also why each step is crucial. So, buckle up, grab your thinking caps, and let's get started!

1. Setting the Stage: The Foundation of Logarithmic Equations

Before we even think about manipulating the equation, we need to understand the fundamental principle that governs logarithmic equations: if $\log_b(A) = \log_b(B)$, then A = B, provided that A and B are positive. This is the bedrock upon which our entire solution rests. It's like the foundation of a building – if it's shaky, the whole structure crumbles. So, let's make sure our foundation is rock solid. Essentially, if two logarithms with the same base are equal, then their arguments (the things inside the logarithm) must also be equal. This allows us to transition from the logarithmic world to the algebraic world, where we can use familiar techniques to solve for our variable, x. But remember, this principle only holds true if both A and B are positive. Why? Because logarithms are only defined for positive arguments. You can't take the logarithm of a negative number or zero – it's simply not allowed in the mathematical universe. This is a crucial detail that we'll need to keep in mind later when we check our solutions. We need to ensure that our potential solutions don't lead to us taking the logarithm of a negative number or zero in the original equation. Now, with this powerful principle in our arsenal, we're ready to take the first step in solving our equation.

2. From Logarithms to Algebra: Eliminating the Logs

Now, let's apply the principle we just discussed to our equation: $\log (x^2-15)=\log (2 x)$. Notice that we have logarithms on both sides with the same base (since no base is explicitly written, we assume it's base 10). This is perfect! We can directly apply our principle and equate the arguments: $x^2-15=2 x$. Boom! We've successfully made the leap from the logarithmic realm to the algebraic realm. We've transformed a potentially intimidating logarithmic equation into a friendly quadratic equation, something we're much more familiar with. This step is like translating a sentence from a foreign language into your native tongue. Once you understand the language, the problem becomes much clearer. Now, we have a quadratic equation staring us in the face. What's the next move? Well, we want to solve for x, and quadratic equations have a special way of being solved – by setting them equal to zero. This is because the standard techniques for solving quadratic equations, such as factoring or the quadratic formula, rely on having the equation in the form ax² + bx + c = 0. So, let's get our equation into that form.

3. Taming the Quadratic: Setting the Stage for Factoring

Our next goal is to rearrange the equation $x^2-15=2 x$ into the standard quadratic form. To do this, we need to move all the terms to one side, leaving zero on the other. The most natural way to do this is to subtract 2x from both sides. This gives us: $x^2-2 x-15=0$. Ta-da! We now have a classic quadratic equation ready to be solved. This step is like organizing your tools before starting a project. You need everything in its place so you can work efficiently. Now, we have a quadratic equation in the standard form. What's the best way to solve it? There are a couple of options: we could use the quadratic formula, which is a guaranteed method for finding the solutions to any quadratic equation. However, before we resort to the formula, it's always a good idea to check if the equation can be factored. Factoring is often a quicker and simpler method, if it's possible. So, let's see if we can factor this quadratic.

4. Cracking the Code: Factoring the Quadratic

Now comes the fun part: factoring the quadratic $x^2-2 x-15=0$. We need to find two numbers that multiply to -15 and add up to -2. Think of it as a puzzle – we're trying to find the missing pieces that fit together perfectly. After a little thought, we can see that the numbers -5 and 3 satisfy these conditions: (-5) * 3 = -15 and (-5) + 3 = -2. So, we can factor the quadratic as follows: $(x-5)(x+3)=0$. We've successfully factored the quadratic! This step is like finding the secret code that unlocks a door. Once you have the code, you can move forward with ease. Now that we've factored the quadratic, we have a product of two factors that equals zero. This is a crucial point because of the zero product property, which states that if the product of two factors is zero, then at least one of the factors must be zero. This property is the key to finding the solutions to our equation.

5. Unveiling the Solutions: The Potential Answers

Now that we have the factored equation $(x-5)(x+3)=0$, we can use the zero product property. This property tells us that either (x - 5) = 0 or (x + 3) = 0. Let's solve each of these equations separately. For x - 5 = 0, we add 5 to both sides, giving us $x=5$. For x + 3 = 0, we subtract 3 from both sides, giving us $x=-3$. So, we have two potential solutions: x = 5 and x = -3. It's tempting to circle these answers and move on, but we're not quite done yet! Remember that crucial detail we discussed earlier about logarithms only being defined for positive arguments? We need to check if these potential solutions actually work in the original equation. This is a critical step that many people forget, but it can save you from getting the wrong answer. We need to make sure that plugging these values back into the original equation doesn't result in us taking the logarithm of a negative number or zero. So, let's put on our detective hats and investigate these potential solutions.

6. The Final Verdict: Checking for Extraneous Solutions

This is the make-or-break step: checking our potential solutions. We need to plug x = 5 and x = -3 back into the original equation, $\log (x^2-15)=\log (2 x)$, and see if they make the equation true and, more importantly, if they result in us taking the logarithm of a positive number. Let's start with x = 5. Plugging it into the equation, we get: $\log (5^2-15)=\log (2 * 5)$ which simplifies to $\log (25-15)=\log (10)$ and further to $\log (10)=\log (10)$. This is true, and both arguments are positive, so x = 5 is a valid solution. Hooray! Now, let's check x = -3. Plugging it into the equation, we get: $\log ((-3)^2-15)=\log (2 * -3)$ which simplifies to $\log (9-15)=\log (-6)$ and further to $\log (-6)=\log (-6)$. Uh oh! We have a problem. We're taking the logarithm of a negative number (-6), which is not allowed. This means that x = -3 is not a valid solution. It's an extraneous solution – a solution that we found algebraically but doesn't actually satisfy the original equation. Extraneous solutions can arise when we perform operations that are not reversible, such as squaring both sides of an equation or, in this case, applying the principle that if log(A) = log(B), then A = B, without considering the domain of the logarithmic function. So, after careful checking, we've determined that x = -3 is not a solution. It's a false lead that we've successfully identified and discarded. This step is like the final quality control check in a factory. You need to make sure your product is perfect before you ship it out.

7. The Grand Finale: Presenting the Solution

After our journey through the world of logarithmic equations, we've arrived at our final destination: the solution! We started with a seemingly complex equation, but by breaking it down into manageable steps, we've successfully conquered it. We applied the fundamental principle of logarithms, transformed the equation into a quadratic, factored it, found potential solutions, and, most importantly, checked for extraneous solutions. And after all that hard work, we can confidently say that the only valid solution to the equation $\log (x^2-15)=\log (2 x)$ is x = 5. That's it! We've solved the puzzle. This final step is like presenting your masterpiece to the world. You've put in the effort, and now you can proudly display your result. So, there you have it, guys! We've successfully navigated the world of logarithmic equations and emerged victorious. Remember, the key to solving these types of problems is to break them down into smaller, more manageable steps, understand the underlying principles, and always, always check your solutions. Keep practicing, and you'll become a logarithmic equation-solving pro in no time!

Based on the steps we've discussed, here's the order in which you'd solve the equation:

1. $x^2-15=2 x$ (Apply the principle: If log(A) = log(B), then A = B)
2. $x^2-2 x-15=0$ (Rearrange into standard quadratic form)
3. $x-5=0$ or $x+3=0$ (Factor the quadratic equation)
4. Potential solutions are -3 and 5 (Solve for x)
5. Check for extraneous solutions and present the final answer (x=5)