Solving Logarithmic Equations: Step-by-Step Guide

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Hey guys! Let's break down how to solve logarithmic equations. It might seem tricky at first, but if we follow a step-by-step approach, it becomes super manageable. We'll use the example equation log(x^2 - 15) = log(2x) to guide us through the process. So, grab your thinking caps, and let's get started!

1. Setting Up the Equation: Eliminating Logarithms

The very first thing we need to do when tackling a logarithmic equation is to get rid of those pesky logarithms! Remember, the goal is to isolate 'x', and we can't do that if it's trapped inside a log. In our case, we have logs on both sides of the equation: log(x^2 - 15) = log(2x). Since the logarithms on both sides have the same base (which is base 10, by the way, if no base is explicitly written), we can simply equate the arguments. This means we can drop the logs and set what's inside them equal to each other. So, the first crucial step is to transform our logarithmic equation into a more manageable algebraic equation. In this specific example, we start with log(x^2 - 15) = log(2x). The key here is recognizing that if log_b(A) = log_b(B), then A = B. Applying this principle, we eliminate the logarithms and move to the next step. By dropping the logs, we're left with x^2 - 15 = 2x. This is a quadratic equation, which we know how to solve! This step is fundamental because it simplifies the problem significantly, allowing us to use familiar algebraic techniques to find the solutions. So, to recap, the initial transformation from a logarithmic equation to an algebraic equation is essential for solving logarithmic problems. Without this, we'd be stuck trying to manipulate logs, which isn't as straightforward as dealing with polynomials.

2. Rearranging into a Quadratic Equation: Getting Ready to Factor

Now that we've ditched the logs, we have a regular equation to work with: x^2 - 15 = 2x. But it's not quite in the form we need yet. To solve for 'x', we need to rearrange this into a standard quadratic equation. You might remember the standard form: ax^2 + bx + c = 0. This form is super helpful because it sets us up perfectly for factoring (or using the quadratic formula, if factoring doesn't work). So, how do we get our equation into that form? We need to move all the terms to one side, leaving zero on the other side. In our case, we have x^2 - 15 = 2x. To get it into the standard form, we subtract 2x from both sides. This gives us x^2 - 2x - 15 = 0. Ta-da! We've successfully rearranged the equation into the standard quadratic form. This is a crucial step because it allows us to apply techniques like factoring or using the quadratic formula. Without this rearrangement, it would be much harder to identify the coefficients and find the solutions. By manipulating the equation into this familiar format, we can easily see the quadratic, linear, and constant terms, which is essential for the next steps in solving the equation. Think of it as organizing your toolbox before starting a project; having everything in the right place makes the job much smoother. This rearrangement is a fundamental step in solving quadratic equations and is a skill that will come in handy in various mathematical contexts.

3. Factoring the Quadratic Equation: Finding Potential Solutions

Alright, we've got our quadratic equation in the standard form: x^2 - 2x - 15 = 0. Now comes the fun part – factoring! Factoring is like reverse-multiplying; we're trying to find two binomials that, when multiplied together, give us our quadratic. To factor x^2 - 2x - 15, we need to find two numbers that multiply to -15 (the constant term) and add up to -2 (the coefficient of the x term). Think about it for a sec… What numbers fit the bill? If you're thinking -5 and 3, you're spot on! -5 multiplied by 3 is -15, and -5 plus 3 is -2. So, we can rewrite our quadratic equation as (x - 5)(x + 3) = 0. See how we used those numbers to create the binomials? This step is super important because it breaks down a complex quadratic equation into simpler linear equations. Once we have the equation in factored form, we can use 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 is a powerful tool that allows us to find the possible values of 'x'. By factoring, we've essentially transformed a single quadratic equation into two separate, easier-to-solve equations. This makes finding the potential solutions much more straightforward. The ability to factor quadratics is a fundamental skill in algebra, and it's a key step in solving many different types of problems. It's like having a secret code that unlocks the solutions to the equation.

4. Solving for x: Identifying Potential Roots

Okay, we've factored the quadratic equation into (x - 5)(x + 3) = 0. Now it's time to unleash the power of the zero-product property! This property basically says that if two things multiplied together equal zero, then at least one of them has to be zero. So, either (x - 5) = 0 or (x + 3) = 0. This gives us two simple little equations to solve. Let's tackle them one at a time. If x - 5 = 0, we just add 5 to both sides, and we get x = 5. Bam! One potential solution. Now, let's look at the other one. If x + 3 = 0, we subtract 3 from both sides, and we get x = -3. Another potential solution! So, factoring and applying the zero-product property, we've found two possible answers for 'x': 5 and -3. But hold your horses! We're not quite done yet. These are just potential solutions. We need to make sure they actually work in the original equation, which involves logarithms. This is a crucial step in solving logarithmic equations because logarithms have certain restrictions on their inputs. For example, we can't take the logarithm of a negative number or zero. By solving each of these equations separately, we isolate 'x' and identify the possible values that could make the original equation true. This step is a direct application of the zero-product property, which is a cornerstone of solving factored equations. It’s like finding the keys that might open the door, but we still need to check if they actually fit the lock. So, these potential solutions are exciting, but we have one more important step to complete.

5. Verifying Solutions: The Crucial Check

We've found our potential solutions: x = 5 and x = -3. But remember, these are just potential solutions. With logarithmic equations, it's super important to check if these values actually work in the original equation. Why? Because logarithms have domain restrictions. You can't take the logarithm of a negative number or zero. So, let's go back to our original equation: log(x^2 - 15) = log(2x). We'll plug in each potential solution and see if it makes the equation true. First, let's try x = 5: log(5^2 - 15) = log(2 * 5) log(25 - 15) = log(10) log(10) = log(10) This checks out! So, x = 5 is a valid solution. Now, let's try x = -3: log((-3)^2 - 15) = log(2 * -3) log(9 - 15) = log(-6) log(-6) = log(-6) Uh oh! We're trying to take the logarithm of a negative number (-6), which isn't allowed. So, x = -3 is not a valid solution. It's called an extraneous solution – a value that we got during the solving process, but it doesn't actually satisfy the original equation. This step is absolutely essential because it ensures that our solutions are valid within the context of the original logarithmic equation. It’s like double-checking your work to make sure you haven’t made any mistakes. By verifying our solutions, we eliminate any extraneous roots and arrive at the correct answer. Therefore, the only valid solution to the equation log(x^2 - 15) = log(2x) is x = 5. This final check is what separates a good solution from a potentially incorrect one.

So, to recap, the correct order of steps is:

  1. x^2 - 15 = 2x (Eliminating the logarithms)
  2. x^2 - 2x - 15 = 0 (Rearranging into a quadratic equation)
  3. (x - 5)(x + 3) = 0 (Factoring the quadratic equation)
  4. x - 5 = 0 or x + 3 = 0 (Solving for x and identifying potential roots)
  5. Potential solutions are -3 and 5 (Verifying solutions and identifying extraneous roots)

And that's how you solve logarithmic equations, guys! Remember to take it one step at a time, and don't forget that crucial verification step at the end. You got this!