Solving Logarithmic Equations A Step-by-Step Guide

Hey guys! Let's dive into solving a logarithmic equation today. We've got a fun one here: log₃(x² - 8) = log₃(2x). This might look intimidating at first, but don't worry, we'll break it down step by step and make it super easy to understand. Our main goal is to find the values of 'x' that satisfy this equation. So, buckle up, and let's get started!

Understanding Logarithmic Equations

Before we jump into the solution, it's essential to understand what logarithmic equations are all about. Logarithms are essentially the inverse operation of exponentiation. Think of it like this: if 2³ = 8, then log₂(8) = 3. The logarithm tells you what exponent you need to raise the base (in this case, 2) to, in order to get a specific number (which is 8 here). Logarithmic equations are equations where the variable (like 'x' in our problem) appears inside a logarithm. Solving these equations involves using the properties of logarithms to isolate the variable and find its value. Key properties that we will use include the fact that if logₐ(b) = logₐ(c), then b = c, and the understanding that the argument of a logarithm (the part inside the parenthesis) must be positive. This is because you can't take the logarithm of a negative number or zero. Understanding these basics will make solving the equation log₃(x² - 8) = log₃(2x) a breeze!

Step-by-Step Solution

Step 1: Equate the Arguments

The beauty of this equation lies in its structure. We have logarithms with the same base (3) on both sides of the equation. This allows us to use a crucial property of logarithms: if logₐ(b) = logₐ(c), then b = c. In our case, this means if log₃(x² - 8) = log₃(2x), then we can safely say that x² - 8 = 2x. This step simplifies the equation dramatically by removing the logarithms and leaving us with a standard algebraic equation to solve. So, by equating the arguments, we've transformed a complex-looking logarithmic equation into a much more manageable quadratic equation. This is a common and powerful technique in solving logarithmic equations, so keep it in mind for future problems!

Step 2: Rearrange into a Quadratic Equation

Now that we have x² - 8 = 2x, we need to rearrange this into the standard form of a quadratic equation, which is ax² + bx + c = 0. This form makes it easier to solve using various methods like factoring, completing the square, or the quadratic formula. To rearrange, we simply subtract 2x from both sides of the equation. This gives us x² - 2x - 8 = 0. See how neatly that fits the standard quadratic form? The 'a' coefficient is 1, 'b' is -2, and 'c' is -8. This rearrangement is a key step because it allows us to apply well-known techniques for solving quadratic equations. Without it, we'd be stuck! So, always aim to get your equation into this form when dealing with quadratic expressions.

Step 3: Solve the Quadratic Equation

We've got our quadratic equation: x² - 2x - 8 = 0. There are a few ways to solve this, but factoring is often the quickest if it's possible. We need to find two numbers that multiply to -8 and add up to -2. After a little thought, we can see that -4 and 2 fit the bill perfectly (-4 * 2 = -8 and -4 + 2 = -2). So, we can factor the quadratic equation as (x - 4)(x + 2) = 0. Now, the zero-product property comes into play: if the product of two factors is zero, then at least one of the factors must be zero. This means either x - 4 = 0 or x + 2 = 0. Solving these simple equations gives us two potential solutions: x = 4 and x = -2. However, we're not done yet! We need to check these solutions in the original logarithmic equation to make sure they're valid.

Step 4: Check for Extraneous Solutions

This is a crucial step that many people forget, but it's super important when dealing with logarithmic equations. Remember, the argument of a logarithm (the part inside the parenthesis) must be positive. We found two potential solutions: x = 4 and x = -2. Let's plug them back into the original equation, log₃(x² - 8) = log₃(2x), and see what happens.

• For x = 4:
  • x² - 8 = 4² - 8 = 16 - 8 = 8 (which is positive)
  • 2x = 2 * 4 = 8 (which is also positive)
  • So, x = 4 is a valid solution.
• For x = -2:
  • x² - 8 = (-2)² - 8 = 4 - 8 = -4 (which is negative!)
  • 2x = 2 * -2 = -4 (also negative!)
  • Since we can't take the logarithm of a negative number, x = -2 is an extraneous solution. It's a solution we found algebraically, but it doesn't actually work in the original equation.

This checking step is vital because logarithmic functions are only defined for positive arguments. Extraneous solutions can creep in due to the nature of logarithmic transformations, so always double-check your answers!

Final Answer

After carefully working through each step, we've arrived at the final answer. We started with the logarithmic equation log₃(x² - 8) = log₃(2x), simplified it, solved the resulting quadratic equation, and then, most importantly, checked for extraneous solutions. We found two potential solutions, x = 4 and x = -2, but after checking, we discovered that x = -2 is an extraneous solution. Therefore, the only valid solution to the equation is x = 4. This highlights the importance of not just solving the equation but also verifying the solutions in the original context.

Practice Problems

To really nail down your understanding of solving logarithmic equations, here are a few practice problems you can try:

1. log₂(x² - 3) = log₂(2x)
2. log₅(x² + 6) = log₅(5x)
3. log₄(x² - 4) = log₄(3x)

Work through these step-by-step, remembering to check for extraneous solutions at the end. Solving these will give you confidence in your ability to tackle logarithmic equations. Happy solving!

Conclusion

Solving logarithmic equations might seem tricky at first, but as we've seen with the equation log₃(x² - 8) = log₃(2x), it's totally manageable with a step-by-step approach. The key is to understand the properties of logarithms, especially the one that allows us to equate arguments when the bases are the same. Rearranging the equation into a standard quadratic form, factoring (or using the quadratic formula), and, crucially, checking for extraneous solutions are all essential steps. Remember, extraneous solutions can pop up because the argument of a logarithm must be positive. By diligently checking our solutions, we ensure that we only accept the valid ones. With practice, these steps become second nature, and you'll be solving logarithmic equations like a pro! So keep practicing, and don't be afraid to tackle those challenging problems.