Solving The Logarithmic Equation 5 - Log₁₀√(1+x) + 4log₁₀√(1-x) = Log₁₀(1/√(1-x²))

Introduction

In the realm of mathematics, logarithmic equations often present intriguing challenges that require a blend of algebraic manipulation and a firm understanding of logarithmic properties. The equation at hand, 5 - log₁₀√(1+x) + 4log₁₀√(1-x) = log₁₀(1/√(1-x²)), is a prime example of such a challenge. This problem invites us to delve into the world of logarithms, unraveling its complexities to arrive at a solution for x. Our goal is not just to find the value of x, but to ultimately determine the value of 100x, adding an extra layer of intrigue to the problem.

Before we embark on the journey of solving this equation, it's crucial to lay the groundwork by understanding the fundamental properties of logarithms. These properties serve as the bedrock upon which our solution will be built. We'll need to recall the power rule, the product rule, the quotient rule, and the change of base formula, among others. These tools will enable us to simplify the equation, combine logarithmic terms, and eventually isolate x. Moreover, we must be mindful of the domain of logarithmic functions, ensuring that our solutions are valid within the context of the original equation. This involves checking for extraneous solutions that may arise during the algebraic manipulation process.

The problem not only tests our proficiency in logarithmic manipulation but also our ability to recognize and apply algebraic identities. The expression 1 - x² lurking within the equation hints at the potential use of the difference of squares factorization, a common yet powerful technique in algebra. By recognizing such patterns, we can often simplify complex expressions into more manageable forms, paving the way for a solution. This interplay between logarithmic properties and algebraic techniques is what makes this problem both challenging and rewarding.

In this article, we will meticulously dissect the equation, step by step, employing a combination of logarithmic identities and algebraic manipulations. We will transform the equation into a more tractable form, gradually revealing the value of x. Furthermore, we will emphasize the importance of checking for extraneous solutions, a crucial step in ensuring the validity of our results. By the end of this exploration, we aim to not only solve the equation but also to provide a clear and comprehensive understanding of the underlying concepts and techniques involved. So, let's embark on this mathematical journey and unravel the solution to this intriguing logarithmic puzzle.

Dissecting the Logarithmic Equation

Our initial equation, 5 - log₁₀√(1+x) + 4log₁₀√(1-x) = log₁₀(1/√(1-x²)), presents a fascinating challenge that requires a strategic approach. To begin, let's break down the equation into smaller, more manageable parts. Our first task is to simplify the logarithmic terms, leveraging the power rule of logarithms. This rule states that logₐ(bⁿ) = nlogₐ(b), allowing us to move exponents outside the logarithm, thereby simplifying the expressions.

Applying the power rule, we can rewrite log₁₀√(1+x) as (1/2)log₁₀(1+x) and log₁₀√(1-x) as (1/2)log₁₀(1-x). Similarly, the term log₁₀(1/√(1-x²)) can be rewritten as log₁₀((1-x²)^(-1/2)), which further simplifies to (-1/2)log₁₀(1-x²). These transformations are crucial as they allow us to work with simpler logarithmic expressions, making the equation less daunting.

Now, our equation looks like this: 5 - (1/2)log₁₀(1+x) + 4(1/2)log₁₀(1-x) = (-1/2)log₁₀(1-x²). Notice that we have a common factor of (1/2) in several terms, which suggests a potential simplification. Multiplying the entire equation by 2 will eliminate this fraction, making the coefficients cleaner and the equation easier to manipulate. This step showcases a common yet effective technique in solving equations: eliminating fractions to reduce complexity.

After multiplying by 2, our equation transforms into 10 - log₁₀(1+x) + 4log₁₀(1-x) = -log₁₀(1-x²). We're making progress! The equation is now free of fractional coefficients, and we can focus on combining the logarithmic terms. This is where the product and quotient rules of logarithms come into play. These rules allow us to combine multiple logarithmic terms into a single logarithm, a powerful technique for simplifying logarithmic equations. The product rule states that logₐ(b) + logₐ(c) = logₐ(bc), while the quotient rule states that logₐ(b) - logₐ(c) = logₐ(b/c). By strategically applying these rules, we can condense the logarithmic terms and move closer to isolating x.

However, before we can apply these rules, we need to address the constant term, 10, on the left side of the equation. To effectively combine it with the logarithmic terms, we must express it as a logarithm with the same base, 10. This is where the property logₐ(aⁿ) = n comes into play. We can rewrite 10 as log₁₀(10¹⁰), allowing us to seamlessly integrate it into the logarithmic expressions. This step highlights the importance of recognizing and applying logarithmic identities to transform constants into logarithmic form, a crucial skill in solving logarithmic equations.

Applying Logarithmic Properties and Algebraic Manipulation

Having simplified the equation to 10 - log₁₀(1+x) + 4log₁₀(1-x) = -log₁₀(1-x²), our next strategic move is to express the constant term, 10, as a logarithm with base 10. As we discussed earlier, 10 can be rewritten as log₁₀(10¹⁰). This transformation is crucial because it allows us to combine all terms into a single logarithmic expression, paving the way for further simplification. By expressing the constant term in logarithmic form, we bridge the gap between the constant and logarithmic terms, enabling us to apply the product and quotient rules effectively.

Substituting log₁₀(10¹⁰) for 10, our equation now reads: log₁₀(10¹⁰) - log₁₀(1+x) + 4log₁₀(1-x) = -log₁₀(1-x²). The stage is set for the application of the logarithmic rules. The next step involves simplifying the term 4log₁₀(1-x). Using the power rule of logarithms, which states that nlogₐ(b) = logₐ(bⁿ), we can rewrite 4log₁₀(1-x) as log₁₀((1-x)⁴). This transformation is essential as it allows us to combine this term with the other logarithmic terms using the product and quotient rules.

Now, our equation takes the form: log₁₀(10¹⁰) - log₁₀(1+x) + log₁₀((1-x)⁴) = -log₁₀(1-x²). We're ready to unleash the power of the product and quotient rules. Recall that the product rule states logₐ(b) + logₐ(c) = logₐ(bc), and the quotient rule states logₐ(b) - logₐ(c) = logₐ(b/c). Applying these rules strategically, we can combine the logarithmic terms on the left side of the equation into a single logarithm. Specifically, we can combine the terms as follows: log₁₀(10¹⁰) - log₁₀(1+x) + log₁₀((1-x)⁴) = log₁₀((10¹⁰ * (1-x)⁴) / (1+x)). This step demonstrates the elegance and efficiency of logarithmic rules in simplifying complex expressions.

Our equation is now significantly more streamlined: log₁₀((10¹⁰ * (1-x)⁴) / (1+x)) = -log₁₀(1-x²). The next step involves addressing the negative sign on the right side of the equation. Recall that -logₐ(b) can be rewritten as logₐ(b⁻¹) or logₐ(1/b). Applying this property, we can transform -log₁₀(1-x²) into log₁₀(1/(1-x²)). This transformation is crucial as it allows us to equate the arguments of the logarithms on both sides of the equation.

With this transformation, our equation becomes: log₁₀((10¹⁰ * (1-x)⁴) / (1+x)) = log₁₀(1/(1-x²)). We've reached a pivotal point in our solution. Since the logarithms on both sides of the equation have the same base, we can equate their arguments. This means that (10¹⁰ * (1-x)⁴) / (1+x) = 1/(1-x²). This step is a cornerstone of solving logarithmic equations: once the equation is reduced to the form logₐ(b) = logₐ(c), we can confidently equate b and c. We've successfully eliminated the logarithms and are now dealing with a purely algebraic equation.

Solving the Algebraic Equation and Finding 100x

Having equated the arguments of the logarithms, we now have the algebraic equation (10¹⁰ * (1-x)⁴) / (1+x) = 1/(1-x²). Our next task is to solve for x. This involves a series of algebraic manipulations, including cross-multiplication and simplification. The first step is to cross-multiply, which gives us 10¹⁰ * (1-x)⁴ * (1-x²) = 1 + x. This maneuver clears the fractions, making the equation more manageable.

Now, we can recognize that (1-x²) can be factored as (1-x)(1+x) using the difference of squares identity. Substituting this factorization into our equation, we get 10¹⁰ * (1-x)⁴ * (1-x)(1+x) = 1 + x. This factorization is a key step as it introduces a common factor of (1+x) on both sides of the equation, which we can potentially cancel out. However, we must be cautious when canceling factors, as we need to consider the possibility that the canceled factor could be zero, which would lead to extraneous solutions.

Before canceling (1+x), let's consider the case where 1+x = 0. This gives us x = -1. We must check if this value satisfies the original equation. Substituting x = -1 into the original equation, we find that the terms log₁₀√(1+x) and log₁₀(1/√(1-x²)) are undefined because they involve taking the logarithm of zero. Therefore, x = -1 is an extraneous solution and must be discarded. This highlights the crucial step of checking for extraneous solutions in logarithmic equations.

Now that we've addressed the potential extraneous solution, we can safely cancel the (1+x) factor from both sides of the equation, provided that x ≠ -1. This gives us 10¹⁰ * (1-x)⁴ * (1-x) = 1, which simplifies to 10¹⁰ * (1-x)⁵ = 1. We're making significant progress! The equation is becoming more streamlined, and we're closer to isolating x.

To further isolate x, we can divide both sides of the equation by 10¹⁰, resulting in (1-x)⁵ = 10⁻¹⁰. Next, we take the fifth root of both sides to eliminate the exponent, giving us 1-x = (10⁻¹⁰)^(1/5) = 10⁻². This step showcases the importance of understanding exponential and radical operations in solving equations. We've successfully removed the exponent and are now dealing with a much simpler equation.

From 1-x = 10⁻², we can easily solve for x. Adding x and subtracting 10⁻² from both sides, we get x = 1 - 10⁻² = 1 - 0.01 = 0.99. We've finally found the value of x! However, our ultimate goal is to find the value of 100x. Multiplying our value of x by 100, we get 100x = 100 * 0.99 = 99. Therefore, the solution to the problem is 100x = 99.

Conclusion

In conclusion, we have successfully navigated the intricate landscape of the logarithmic equation 5 - log₁₀√(1+x) + 4log₁₀√(1-x) = log₁₀(1/√(1-x²)) to arrive at the solution 100x = 99. This journey has showcased the power and elegance of logarithmic properties, algebraic manipulation, and careful consideration of extraneous solutions. We began by dissecting the equation, applying the power rule of logarithms to simplify individual terms. We then strategically used the product and quotient rules to combine multiple logarithmic terms into a single, more manageable expression. The crucial step of expressing the constant term as a logarithm with the same base allowed us to seamlessly integrate it into the equation.

Throughout the solution process, we emphasized the importance of algebraic techniques, such as factoring and cross-multiplication, in simplifying the equation and isolating x. The recognition of the difference of squares identity proved to be a pivotal step in our simplification strategy. However, we also highlighted the potential pitfalls of algebraic manipulation, particularly the introduction of extraneous solutions. The meticulous step of checking for extraneous solutions, especially after canceling common factors, ensured the validity of our final answer.

Our journey through this logarithmic equation serves as a testament to the interconnectedness of mathematical concepts. The solution required a harmonious blend of logarithmic identities, algebraic techniques, and a keen awareness of potential pitfalls. By carefully applying these tools, we not only found the value of x but also gained a deeper appreciation for the beauty and power of mathematics. The final answer, 100x = 99, stands as a testament to our perseverance and the effectiveness of our problem-solving strategies.

This exploration also underscores the importance of a systematic approach to problem-solving. Breaking down complex problems into smaller, more manageable steps, carefully applying relevant rules and techniques, and diligently checking for extraneous solutions are all essential components of successful problem-solving in mathematics. By adopting such a methodical approach, we can confidently tackle even the most challenging mathematical puzzles.