Logarithmic Identity Proof: A Detailed Mathematical Solution

Hey guys! Today, we're diving deep into a fascinating problem involving logarithms. This isn't your everyday math question; it requires a solid understanding of logarithmic properties and a bit of algebraic manipulation. So, buckle up and let's get started! We're going to break down this problem step-by-step, making sure everyone can follow along. Our main goal is to prove a specific logarithmic identity. Trust me; it’s going to be an enlightening journey. Let's dive into the heart of this problem. The first key to solving this logarithmic puzzle lies in understanding the change of base formula. This formula is a game-changer when dealing with logarithms of different bases. It allows us to convert logarithms from one base to another, which is exactly what we need in this scenario. Remember, the change of base formula is your best friend when logarithms get tricky. Keep this formula at the forefront of your mind as we progress through the solution. It will be our guiding light, ensuring we stay on the right track. Think of the change of base formula as a universal translator for logarithms, enabling us to speak fluently in the language of any base. Without it, our task would be significantly more challenging, if not impossible. Now, with the change of base formula firmly in our toolkit, we're ready to tackle the problem head-on. Let’s get to work!

Problem Statement

Before we jump into the solution, let's clearly state the problem we're tackling. Given that x ∈ (0,1) ∪ (1, +∞), and if lg x = a, log_6 x = b, and log_{15} x = c, we need to prove that:

1/log_2 x + 1/log_3 x + 1/log_5 x = 1/2 (1/a + 1/b + 1/c)

This equation looks a bit intimidating at first glance, but don't worry, we'll break it down into manageable pieces. The key here is to see the relationships between the different logarithmic expressions. Notice how we have logarithms with different bases (2, 3, 5, 6, and 15), and we also have the common logarithm (base 10), represented as lg x. Our strategy will be to use the properties of logarithms, particularly the change of base formula, to express everything in terms of a common base. This will allow us to simplify the equation and eventually prove the identity. Remember, in mathematics, clarity is key. A well-defined problem is half the solution. So, let’s keep this statement in mind as we move forward, ensuring we stay focused on our ultimate goal. This is the roadmap for our journey, guiding us toward the final destination of proving this identity. Let's move on to the next step, where we start unraveling the complexities of this equation.

Utilizing Logarithmic Properties

Our first step in solving this problem involves leveraging the fundamental properties of logarithms. Specifically, we'll be using the change of base formula, which, as we mentioned earlier, is crucial for handling logarithms with different bases. This property is what allows us to bridge the gap between the different logarithmic expressions in our equation. The change of base formula states that log_b a = (log_k a) / (log_k b), where k can be any base. We'll strategically use this formula to convert all logarithms to a common base, which will make the subsequent simplification process much easier. Think of it as translating different languages into a common tongue, allowing us to understand and manipulate them effectively. This step is not just about applying a formula; it's about choosing the right tool for the job and understanding why it works. By converting to a common base, we're essentially leveling the playing field, making the logarithmic expressions comparable and workable. This is a crucial step in our journey to prove the identity. Let's see how this powerful tool helps us unravel the problem.

Applying the Change of Base Formula

Let’s start by applying the change of base formula to the left-hand side of the equation. We have 1/log_2 x, 1/log_3 x, and 1/log_5 x. A useful trick here is to remember that 1/log_b a = log_a b. This is a direct consequence of the change of base formula and simplifies our work significantly. Applying this, we get:

• 1/log_2 x = log_x 2
• 1/log_3 x = log_x 3
• 1/log_5 x = log_x 5

So, the left-hand side of our equation now looks like this: log_x 2 + log_x 3 + log_x 5. Isn't that much cleaner already? We've transformed the reciprocals of logarithms into simpler logarithmic expressions. This is a perfect example of how understanding logarithmic properties can drastically simplify complex expressions. Now, we can use another important property of logarithms: log_b m + log_b n = log_b (mn). This property allows us to combine the sum of logarithms into a single logarithm. By combining these logarithms, we are effectively condensing the expression, bringing us closer to a more manageable form. Remember, the goal is to simplify the equation step by step, making it easier to manipulate and eventually prove the identity. Let’s apply this property in the next step and see how much further we can simplify the left-hand side.

Combining Logarithms

Using the property log_b m + log_b n = log_b (mn), we can combine the terms on the left-hand side:

log_x 2 + log_x 3 + log_x 5 = log_x (2 * 3 * 5) = log_x 30

So, the left-hand side of our equation has beautifully simplified to log_x 30. We've taken three separate logarithmic terms and condensed them into a single, elegant expression. This is a testament to the power of logarithmic properties and how they can be used to manipulate and simplify complex equations. Now, let’s shift our focus to the right-hand side of the equation. We need to work our magic there as well, aiming to express it in a form that we can easily compare with the simplified left-hand side. This is where our initial conditions come into play. We'll need to use the given values of a, b, and c to unravel the complexities of the right-hand side. Remember, the key to solving any mathematical problem is to break it down into smaller, manageable steps. We've successfully simplified the left-hand side; now it's time to tackle the right-hand side with the same methodical approach. Let’s dive in!

Simplifying the Right-Hand Side

Now, let’s tackle the right-hand side of the equation: 1/2 (1/a + 1/b + 1/c). Remember that a = lg x = log_{10} x, b = log_6 x, and c = log_{15} x. Our goal here is to express this side in a form that we can compare with the simplified left-hand side, which is log_x 30. The first step is to substitute the values of a, b, and c into the expression. This will give us a clearer picture of what we're working with and pave the way for further simplification. Think of it as replacing the variables with their actual values, making the expression more concrete and easier to manipulate. This is a standard technique in mathematics – substituting known values to simplify equations. By making these substitutions, we're essentially translating the expression into a language we can understand and work with. So, let’s go ahead and make these substitutions and see what we get. It's like putting the pieces of a puzzle together; each substitution brings us closer to the complete picture. Let’s get to it!

Substituting Values

Substituting the values of a, b, and c, we get:

1/2 (1/a + 1/b + 1/c) = 1/2 (1/log_{10} x + 1/log_6 x + 1/log_{15} x)

Now, we have an expression that involves logarithms with different bases. Just like we did on the left-hand side, we can use the property 1/log_b a = log_a b to simplify this further. This property, as we’ve seen, is a powerful tool for dealing with reciprocals of logarithms. By applying it here, we're essentially flipping the fractions, making the expression more amenable to further simplification. This is a strategic move, aiming to bring all the logarithmic terms into a common format. Think of it as rearranging the pieces of a puzzle to better fit together. Each step we take is a deliberate effort to transform the expression into a more manageable form. So, let’s apply this property and see what the right-hand side looks like now. It's all about transforming the complex into the simple, one step at a time. Let’s continue this journey of simplification.

Applying the Reciprocal Property

Applying the property 1/log_b a = log_a b, we get:

1/2 (log_x 10 + log_x 6 + log_x 15)

We're making excellent progress! The expression is starting to look much cleaner. Now, we can use the property log_b m + log_b n = log_b (mn) again to combine the logarithms inside the parentheses. This is the same property we used on the left-hand side, and it's just as effective here. By combining these logarithmic terms, we're condensing the expression, making it more compact and easier to handle. Think of it as merging several small streams into a single, powerful river. Each step we take brings us closer to our goal of proving the identity. So, let’s go ahead and combine these logarithms and see how the expression transforms. It’s like watching a mathematical puzzle come together, piece by piece. Let’s proceed with this exciting simplification process.

Combining Logarithms (Again!)

Using the property log_b m + log_b n = log_b (mn), we combine the logarithms:

1/2 (log_x 10 + log_x 6 + log_x 15) = 1/2 (log_x (10 * 6 * 15))

Now, let's multiply the numbers inside the logarithm:

10 * 6 * 15 = 900

So, our expression becomes:

1/2 (log_x 900)

We’re getting closer! The right-hand side is looking increasingly similar to our simplified left-hand side. Now, we need to express 900 in terms of its prime factors, which will help us further simplify the expression. This is a crucial step, as it allows us to break down a large number into its fundamental building blocks, making it easier to work with. Think of it as dissecting a complex object to understand its individual components. By factoring 900, we're uncovering its hidden structure, which will ultimately help us connect the right-hand side to the left-hand side. So, let’s factor 900 and see what we find. It's like peeling away the layers of an onion, revealing the core underneath. Let’s proceed with this crucial step.

Factoring and Simplifying

Let's factor 900:

900 = 2^2 * 3^2 * 5^2

So, we can rewrite our expression as:

1/2 (log_x (2^2 * 3^2 * 5^2))

Using the property log_b (m^p) = p log_b m, we can further simplify:

1/2 (log_x (2^2 * 3^2 * 5^2)) = 1/2 (log_x ((2 * 3 * 5)^2)) = 1/2 (log_x (30^2))

Applying the same property again:

1/2 (log_x (30^2)) = 1/2 * 2 * log_x 30 = log_x 30

And there you have it! The right-hand side has been simplified to log_x 30, which is exactly what we obtained for the left-hand side. We’ve successfully transformed the right-hand side, step by step, using the fundamental properties of logarithms. This is a testament to the power of consistent application of mathematical principles. We've taken a seemingly complex expression and, through methodical simplification, revealed its underlying simplicity. Now, we’re at the grand finale of our proof. Let’s put it all together and celebrate our achievement!

Conclusion: The Proof

We've shown that:

• Left-hand side: 1/log_2 x + 1/log_3 x + 1/log_5 x = log_x 30
• Right-hand side: 1/2 (1/a + 1/b + 1/c) = log_x 30

Therefore, we have proven that:

1/log_2 x + 1/log_3 x + 1/log_5 x = 1/2 (1/a + 1/b + 1/c)

This was a challenging problem, guys, but we tackled it head-on and emerged victorious! We started with a complex equation and, using the properties of logarithms, we methodically simplified both sides until they matched. This highlights the beauty and power of mathematics – how seemingly complicated problems can be solved with the right tools and a clear strategy. The key takeaways from this exercise are the importance of understanding logarithmic properties, particularly the change of base formula and the rules for combining logarithms. These are fundamental tools in the mathematician's toolkit, and mastering them can unlock solutions to a wide range of problems. Remember, mathematics is not just about memorizing formulas; it's about understanding the underlying principles and applying them creatively. We hope you found this detailed solution helpful and insightful. Keep practicing, keep exploring, and keep the mathematical fire burning! You've conquered this logarithmic challenge, and there are many more mathematical adventures awaiting you. On to the next problem!