Hey there, math enthusiasts! Today, we're diving into the fascinating world of logarithms to solve an equation. Logarithmic equations might seem daunting at first, but with a step-by-step approach, we can crack them open and find the solutions. In this article, we'll be tackling the equation log₆(x) + log₆(x+5) = 2. We will explore the properties of logarithms and demonstrate how to apply them to solve this particular problem. Whether you're a student brushing up on your algebra skills or a math aficionado looking for a fun challenge, this guide is for you. So, grab your pencil and paper, 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. In simple terms, a logarithm answers the question: "To what power must we raise the base to get a certain number?" For instance, log₁₀(100) = 2 because 10 raised to the power of 2 equals 100. The general form of a logarithmic expression is logₐ(b) = c, which is equivalent to aᶜ = b. Here, 'a' is the base, 'b' is the argument, and 'c' is the logarithm.
When dealing with logarithms, it's crucial to remember a few key properties that will help us simplify and solve equations. One of the most important properties is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Mathematically, this is expressed as logₐ(mn) = logₐ(m) + logₐ(n). This property is going to be our best friend in solving the equation at hand. Another essential property is the power rule, which states that the logarithm of a number raised to a power is equal to the power times the logarithm of the number. In equation form, this is logₐ(mⁿ) = n logₐ(m). Lastly, let's not forget the change of base formula, which allows us to convert logarithms from one base to another. This is particularly useful when we need to evaluate logarithms on a calculator that only has common logarithms (base 10) or natural logarithms (base e). The formula is logₓ(a) = logᵧ(a) / logᵧ(x), where y is the new base. By understanding and applying these basic principles, we set a solid foundation for tackling more complex logarithmic equations. So, keep these rules handy, and let’s move forward with confidence!
Applying Logarithmic Properties to the Equation
Now that we've refreshed our understanding of logarithms, let's get back to our equation: log₆(x) + log₆(x+5) = 2. Our first step is to use the product rule of logarithms to combine the two logarithmic terms on the left side of the equation. Remember, the product rule states that logₐ(m) + logₐ(n) = logₐ(mn). Applying this rule to our equation, we get: log₆(x(x+5)) = 2. This simplifies our equation significantly, turning the sum of two logarithms into a single logarithm.
Next, we need to get rid of the logarithm. To do this, we'll convert the logarithmic equation into its exponential form. Recall that logₐ(b) = c is equivalent to aᶜ = b. Applying this to our equation, where the base 'a' is 6, the argument 'b' is x(x+5), and the logarithm 'c' is 2, we get: 6² = x(x+5). This transformation is crucial because it allows us to work with a more familiar algebraic expression. Now, our equation looks much simpler and easier to handle. We've successfully converted the logarithmic equation into a quadratic equation, which we can solve using standard algebraic techniques.
Before we proceed further, it's important to pause and reflect on the steps we've taken. We started with a logarithmic equation, applied the product rule to combine the logarithms, and then converted the equation into exponential form. This methodical approach is key to solving logarithmic equations effectively. Each step is deliberate and brings us closer to the solution. By breaking down the problem into manageable parts, we avoid getting overwhelmed and increase our chances of success. So, with this transformed equation in hand, let's move on to the next phase: solving the resulting quadratic equation.
Solving the Quadratic Equation
We've transformed our logarithmic equation into a quadratic equation: 6² = x(x+5). Let's simplify and solve it. First, we evaluate 6², which gives us 36. So, the equation becomes: 36 = x(x+5). Next, we need to expand the right side of the equation by distributing x across the parentheses: 36 = x² + 5x. Now, to solve this quadratic equation, we need to set it equal to zero. We can do this by subtracting 36 from both sides: x² + 5x - 36 = 0.
Now we have a standard quadratic equation in the form ax² + bx + c = 0, where a = 1, b = 5, and c = -36. There are several ways to solve quadratic equations, including factoring, completing the square, and using the quadratic formula. In this case, factoring seems like the most straightforward approach. We need to find two numbers that multiply to -36 and add up to 5. After some thought, we can see that the numbers 9 and -4 fit the bill since 9 * -4 = -36 and 9 + (-4) = 5. Therefore, we can factor the quadratic equation as follows: (x + 9)(x - 4) = 0.
To find the solutions for x, we set each factor equal to zero: x + 9 = 0 and x - 4 = 0. Solving these linear equations gives us two potential solutions: x = -9 and x = 4. These are the values of x that make our factored equation true. However, we're not done yet! It's crucial to check these solutions in the original logarithmic equation to ensure they are valid. Remember, logarithms have domain restrictions, and we need to make sure that our solutions don't violate these restrictions. Let's move on to the next step: checking our solutions.
Checking for Extraneous Solutions
We've found two potential solutions for our equation: x = -9 and x = 4. However, when dealing with logarithmic equations, it's essential to check for extraneous solutions. These are solutions that satisfy the transformed equation but not the original equation due to domain restrictions of logarithms. Remember, the argument of a logarithm must be positive. In our original equation, log₆(x) + log₆(x+5) = 2, we have two logarithmic terms: log₆(x) and log₆(x+5). This means that both x and x+5 must be greater than zero.
Let's start by checking x = -9. If we substitute -9 for x in the original equation, we get log₆(-9) + log₆(-9+5) = 2. Here, we have log₆(-9) and log₆(-4). Since we cannot take the logarithm of a negative number, x = -9 is an extraneous solution. It does not satisfy the domain restrictions of the logarithmic functions in the original equation. Therefore, we must discard this solution.
Next, let's check x = 4. Substituting 4 for x in the original equation gives us log₆(4) + log₆(4+5) = 2, which simplifies to log₆(4) + log₆(9) = 2. Both arguments, 4 and 9, are positive, so this solution doesn't violate any domain restrictions. To verify that it is a valid solution, we can use a calculator to evaluate the logarithms and check if the equation holds true. We have log₆(4) ≈ 0.7737 and log₆(9) ≈ 1.2263. Adding these together, we get approximately 2, which confirms that x = 4 is indeed a valid solution.
By checking our solutions, we've ensured that we're only accepting values that make sense in the context of the original logarithmic equation. This step is crucial to avoid errors and arrive at the correct answer. So, after careful consideration, we can confidently state that the solution to the equation log₆(x) + log₆(x+5) = 2 is x = 4.
Conclusion: The Solution and Key Takeaways
Alright, guys, we've successfully navigated the world of logarithms and solved the equation log₆(x) + log₆(x+5) = 2! Through a series of steps, we transformed the equation, solved for x, and checked for extraneous solutions. Our journey led us to the final answer: x = 4. This solution satisfies the original equation and respects the domain restrictions of logarithms. Solving logarithmic equations can seem tricky, but with a systematic approach, we can break down the problem into manageable steps and arrive at the correct solution. Remember the key logarithmic properties, such as the product rule, and always check for extraneous solutions to ensure accuracy.
Moreover, this exercise underscores the importance of understanding the fundamental properties of logarithms and how they can be applied to simplify and solve equations. Logarithmic equations appear in various fields, including mathematics, physics, engineering, and computer science. Mastering these concepts will undoubtedly be beneficial in your future academic and professional pursuits. So, keep practicing, stay curious, and continue to explore the fascinating world of mathematics. Whether you're a student, a professional, or simply a math enthusiast, the ability to solve logarithmic equations is a valuable skill that will serve you well. Keep up the great work, and remember, every equation solved is a step forward in your mathematical journey!