Solve Log₆(x) + Log₆(x-1) = 1 A Step-by-Step Guide

Hey everyone! Today, we're diving into the exciting world of logarithmic equations. Logarithmic equations might seem intimidating at first, but don't worry, we'll break it down step by step and make sure you understand exactly how to solve them. We'll be tackling the equation log₆(x) + log₆(x-1) = 1. So, grab your thinking caps, and let's get started!

Understanding Logarithms: The Key to Unlocking the Equation

Before we jump into solving the equation, let's make sure we're all on the same page about what logarithms actually are. At its heart, a logarithm is just the inverse operation of exponentiation. Think of it this way: if 2³ = 8, then log₂(8) = 3. The logarithm (base 2) of 8 is 3 because 2 raised to the power of 3 equals 8. That's the fundamental idea! Remember this key concept because we'll be using it throughout the solution.

More generally, the equation logₐ(b) = c means that aᶜ = b. Here, 'a' is the base of the logarithm, 'b' is the argument (the number we're taking the logarithm of), and 'c' is the exponent. Understanding this relationship is crucial for manipulating logarithmic equations. It's like having the secret code to unlock the problem! You've got to know the base, the argument, and the exponent and how they all connect. For example, log₁₀(100) = 2 because 10² = 100. Or, log₅(25) = 2 because 5² = 25. See the pattern? Keep practicing this conversion between logarithmic and exponential forms, and you'll become a logarithm pro in no time!

Now, why is all this logarithm talk important for our equation, log₆(x) + log₆(x-1) = 1? Well, it's because we need to use the properties of logarithms to simplify the equation. Logarithms have some really cool properties that allow us to combine them, expand them, and generally make them easier to work with. One of the most important properties for this problem is the product rule of logarithms. This rule states that logₐ(m) + logₐ(n) = logₐ(m*n). In other words, the logarithm of the product of two numbers is equal to the sum of the logarithms of those numbers (provided they have the same base). This property is a game-changer for us because it allows us to combine the two logarithms on the left side of our equation into a single logarithm. So, before we start crunching numbers, make sure you've got this product rule locked in your memory. It's going to be our best friend in solving this equation!

Applying the Product Rule: Combining Logarithms

Okay, guys, now that we've refreshed our understanding of logarithms and the product rule, let's apply it to our equation: log₆(x) + log₆(x-1) = 1. Remember that the product rule states logₐ(m) + logₐ(n) = logₐ(m*n). In our case, 'a' is 6, 'm' is x, and 'n' is (x-1). So, we can rewrite the left side of the equation as a single logarithm:

log₆(x) + log₆(x-1) = log₆(x * (x-1))

This is a huge step forward! We've taken two separate logarithms and combined them into one. This simplification makes the equation much easier to handle. Now our equation looks like this:

log₆(x * (x-1)) = 1

Notice how the product rule allowed us to condense the expression. This is a common strategy when solving logarithmic equations: to combine logarithms whenever possible. It reduces the complexity and brings us closer to isolating the variable (in this case, x). So, always be on the lookout for opportunities to use the product rule, the quotient rule (logₐ(m) - logₐ(n) = logₐ(m/n)), or the power rule (logₐ(mᵖ) = p*logₐ(m)) – these are your allies in the fight against complex logarithmic equations!

Now that we have a single logarithm on the left side, we're ready for the next crucial step: converting the logarithmic equation into an exponential equation. This is where that fundamental understanding of the relationship between logarithms and exponents comes into play. Remember, logₐ(b) = c is equivalent to aᶜ = b. We're going to use this equivalence to transform our equation and get rid of that logarithm altogether!

Converting to Exponential Form: Dropping the Logarithm

Alright, let's tackle the conversion to exponential form. We've got log₆(x * (x-1)) = 1. Remember our key relationship: logₐ(b) = c is the same as aᶜ = b. In our equation, 'a' is 6 (the base of the logarithm), 'b' is x * (x-1) (the argument), and 'c' is 1 (the value on the right side of the equation). So, we can rewrite the equation in exponential form as:

6¹ = x * (x-1)

See how the logarithm disappeared? We've effectively "undone" the logarithm by expressing the equation in exponential form. This is a powerful technique! Now we have a much simpler equation to solve. 6¹ is just 6, so we can further simplify the equation to:

6 = x * (x-1)

Now we're dealing with a quadratic equation, which we know how to handle. We're getting closer and closer to finding the value of x! Remember, the goal here was to isolate x, and by converting to exponential form, we've eliminated the logarithm and opened the door to solving for x directly. This step is often the turning point in solving logarithmic equations – it transforms the problem from a logarithmic one into an algebraic one. And once we're in the realm of algebra, we have a whole toolbox of techniques at our disposal.

So, make sure you're comfortable with this conversion process. Practice going back and forth between logarithmic and exponential forms until it feels like second nature. It's a fundamental skill for solving all sorts of logarithmic problems. Now, let's move on to the next step: solving that quadratic equation!

Solving the Quadratic Equation: Finding Potential Solutions

Okay, we've transformed our logarithmic equation into a quadratic equation: 6 = x * (x-1). Let's expand the right side and rearrange the equation to get it into the standard quadratic form (ax² + bx + c = 0):

6 = x² - x

Subtracting 6 from both sides, we get:

x² - x - 6 = 0

Now we have a standard quadratic equation. There are a few ways to solve this. We can try factoring, using the quadratic formula, or even completing the square. In this case, factoring seems like the easiest route. We need to find two numbers that multiply to -6 and add up to -1 (the coefficient of the x term). Those numbers are -3 and 2. So, we can factor the quadratic equation as:

(x - 3)(x + 2) = 0

Now, for the product of two factors to be zero, at least one of them must be zero. This gives us two potential solutions:

x - 3 = 0 or x + 2 = 0

Solving these simple equations, we get:

x = 3 or x = -2

So, we have two potential solutions: x = 3 and x = -2. But hold on! We're not done yet. We need to remember a crucial step when dealing with logarithmic equations: checking for extraneous solutions.

Checking for Extraneous Solutions: The Crucial Verification Step

Alright, we've arrived at two potential solutions for our equation: x = 3 and x = -2. But before we declare victory, we need to perform a very important check: we need to see if these solutions actually work in the original equation. This is because logarithms have a restricted domain – you can only take the logarithm of a positive number. If we plug a solution into the original equation and end up taking the logarithm of a negative number or zero, that solution is called an extraneous solution, and we have to discard it. Think of it like a filter – we need to filter out any solutions that don't make sense in the context of logarithms.

Let's start by checking x = 3. Plugging it into the original equation, log₆(x) + log₆(x-1) = 1, we get:

log₆(3) + log₆(3-1) = log₆(3) + log₆(2)

Both 3 and 2 are positive numbers, so we can take their logarithms. This looks promising! Let's see if it equals 1. Using the product rule, we can combine the logarithms:

log₆(3) + log₆(2) = log₆(3 * 2) = log₆(6)

And what is log₆(6)? Well, 6 raised to what power equals 6? The answer is 1! So, we have:

log₆(6) = 1

Great! x = 3 checks out. It's a valid solution.

Now, let's check x = -2. Plugging it into the original equation, we get:

log₆(-2) + log₆(-2-1) = log₆(-2) + log₆(-3)

Uh oh! We're trying to take the logarithm of -2 and -3, which are negative numbers. This is a big no-no in the world of logarithms. You can't take the logarithm of a negative number or zero. Therefore, x = -2 is an extraneous solution. We have to throw it out!

This step of checking for extraneous solutions is absolutely crucial. Don't skip it! It's the difference between getting the right answer and getting a wrong one. Logarithmic equations are notorious for producing extraneous solutions, so always, always, always check your answers in the original equation.

The Final Answer: The Solution to the Logarithmic Puzzle

We've gone through all the steps: understanding logarithms, applying the product rule, converting to exponential form, solving the quadratic equation, and, most importantly, checking for extraneous solutions. We found two potential solutions: x = 3 and x = -2. But after checking, we discovered that x = -2 is an extraneous solution, and only x = 3 works in the original equation.

Therefore, the solution to the logarithmic equation log₆(x) + log₆(x-1) = 1 is:

x = 3

Congratulations! You've successfully solved a logarithmic equation. Remember, the key is to break it down step by step, understand the properties of logarithms, and always check your answers. Keep practicing, and you'll become a logarithmic equation-solving master!

In summary, to solve logarithmic equations like this one:

1. Understand the basics of logarithms: Know the relationship between logarithms and exponents.
2. Apply logarithmic properties: Use the product rule, quotient rule, and power rule to simplify the equation.
3. Convert to exponential form: Get rid of the logarithm by expressing the equation in exponential form.
4. Solve the resulting equation: This might be a linear, quadratic, or other type of equation.
5. Check for extraneous solutions: This is the most critical step! Plug your potential solutions back into the original equation to make sure they work.

By following these steps, you can confidently tackle any logarithmic equation that comes your way. Keep up the great work, and happy solving!