Solving Logarithmic Equations The Solution To 2 Ln Eln 2x - Ln Eln 10x = Ln 30

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Introduction: Delving into Logarithmic Equations

At the heart of mathematical problem-solving lies the ability to dissect and conquer equations, especially those involving logarithmic functions. Logarithmic equations, while sometimes appearing complex, can be elegantly solved by applying the fundamental properties of logarithms. In this comprehensive exploration, we will dissect the given equation, $2 \ln e^{\ln 2 x}-\ln e^{\ln 10 x}=\ln 30$, step by step, to arrive at the true solution. We will not only pinpoint the correct answer from the provided options but also illuminate the underlying principles that govern logarithmic manipulations. This journey through the equation will serve as a beacon, guiding you through the often-intricate world of logarithmic equations.

Understanding the Landscape of Logarithmic Properties

Before diving headfirst into the equation, it's crucial to arm ourselves with the necessary tools – the properties of logarithms. These properties act as our compass and map, guiding us through the twists and turns of the equation. Key properties that we will leverage include:

1. The Power Rule: ${ \ln a^b = b \ln a }$. This rule allows us to bring exponents down as coefficients, simplifying expressions.
2. The Logarithm of a Product: ${ \ln (ab) = \ln a + \ln b }$. This property enables us to break down the logarithm of a product into the sum of logarithms.
3. The Logarithm of a Quotient: ${ \ln (a/b) = \ln a - \ln b }$. Conversely, this property allows us to express the logarithm of a quotient as the difference of logarithms.
4. The Inverse Relationship: ${ e^{\ln x} = x }$ and ${ \ln e^x = x }$. This fundamental relationship between exponential and logarithmic functions is pivotal in simplifying our equation.

With these properties in our arsenal, we are now well-equipped to embark on our quest to solve the equation.

Step-by-Step Solution: A Journey Through Logarithmic Transformations

Our primary objective is to isolate the variable x by systematically simplifying the equation. Let's embark on this step-by-step journey:

Step 1: Simplifying using the Inverse Relationship

We begin by recognizing the inverse relationship between the natural logarithm and the exponential function. Specifically, we know that ${ e^{\ln x} = x }$ and ${ \ln e^x = x }$. Applying this to our equation, we can simplify the terms within the logarithms:

$2 \ln e^{\ln 2 x}-\ln e^{\ln 10 x}=\ln 30$ becomes:

$2 \ln (2x) - \ln (10x) = \ln 30$

This initial simplification significantly reduces the complexity of the equation, making it more amenable to further manipulation.

Step 2: Applying the Power Rule of Logarithms

Next, we invoke the power rule of logarithms, which states that ${ \ln a^b = b \ln a }$. This rule allows us to move the coefficient in front of the first logarithm into the exponent:

$2 \ln (2x)$ transforms into $\ln (2x)^2$, giving us:

$\ln (2x)^2 - \ln (10x) = \ln 30$

Further simplifying the squared term, we get:

$\ln (4x^2) - \ln (10x) = \ln 30$

This transformation sets the stage for the application of another crucial logarithmic property.

Step 3: Utilizing the Quotient Rule of Logarithms

Now, we employ the quotient rule of logarithms, which states that ${ \ln (a/b) = \ln a - \ln b }$. This property allows us to combine the two logarithms on the left-hand side of the equation:

$\ln (4x^2) - \ln (10x)$ becomes $\ln \left( \frac{4x^2}{10x} \right)$, leading to:

$\ln \left( \frac{4x^2}{10x} \right) = \ln 30$

Simplifying the fraction inside the logarithm, we have:

$\ln \left( \frac{2x}{5} \right) = \ln 30$

The equation is now in a much cleaner form, making the next step more intuitive.

Step 4: Eliminating the Logarithms

Since we have a logarithm on both sides of the equation, we can eliminate them by recognizing that if ${ \ln a = \ln b }$, then ${ a = b }$. Applying this principle, we get:

$\frac{2x}{5} = 30$

This step brings us closer to isolating x and unveiling the solution.

Step 5: Solving for x

To isolate x, we multiply both sides of the equation by 5:

$2x = 30 \times 5$

$2x = 150$

Finally, we divide both sides by 2:

$x = \frac{150}{2}$

$x = 75$

Thus, we have arrived at the solution: x = 75.

Conclusion: The True Solution Unveiled

Through a meticulous application of logarithmic properties and algebraic manipulations, we have successfully navigated the equation $2 \ln e^{\ln 2 x}-\ln e^{\ln 10 x}=\ln 30$ and determined that the true solution is x = 75. This exercise not only provides the answer but also underscores the importance of mastering logarithmic properties in tackling mathematical challenges. Understanding the inverse relationship, power rule, and quotient rule of logarithms empowers us to simplify complex equations and reveal their hidden solutions. Therefore, the correct answer is:

• B. x = 75

This exploration serves as a testament to the power of mathematical principles in unraveling seemingly intricate problems. As we journey further into the world of mathematics, these principles will continue to serve as our guiding stars, illuminating the path to solutions and deeper understanding.

Why Other Options are Incorrect: A Critical Examination

While we've established that x = 75 is the correct solution, it's crucial to understand why the other options are incorrect. This critical examination reinforces our understanding of the problem and the solution process. Let's analyze each incorrect option:

Option A: x = 30

If we substitute x = 30 into the original equation, we get:

$2 \ln e^{\ln (2 \cdot 30)} - \ln e^{\ln (10 \cdot 30)} = \ln 30$

Simplifying, we have:

$2 \ln (60) - \ln (300) = \ln 30$

Using logarithmic properties, this becomes:

$\ln (60^2) - \ln (300) = \ln 30$

$\ln (3600) - \ln (300) = \ln 30$

$\ln \left( \frac{3600}{300} \right) = \ln 30$

$\ln (12) = \ln 30$

This statement is false, as ${ \ln 12 }$ does not equal ${ \ln 30 }$. Therefore, x = 30 is not the solution.

Option C: x = 150

Substituting x = 150 into the original equation:

$2 \ln e^{\ln (2 \cdot 150)} - \ln e^{\ln (10 \cdot 150)} = \ln 30$

Simplifying:

$2 \ln (300) - \ln (1500) = \ln 30$

$\ln (300^2) - \ln (1500) = \ln 30$

$\ln (90000) - \ln (1500) = \ln 30$

$\ln \left( \frac{90000}{1500} \right) = \ln 30$

$\ln (60) = \ln 30$

This statement is also false, as ${ \ln 60 }$ does not equal ${ \ln 30 }$. Hence, x = 150 is not the correct solution.

Option D: x = 300

Substituting x = 300 into the equation:

$2 \ln e^{\ln (2 \cdot 300)} - \ln e^{\ln (10 \cdot 300)} = \ln 30$

Simplifying:

$2 \ln (600) - \ln (3000) = \ln 30$

$\ln (600^2) - \ln (3000) = \ln 30$

$\ln (360000) - \ln (3000) = \ln 30$

$\ln \left( \frac{360000}{3000} \right) = \ln 30$

$\ln (120) = \ln 30$

This is another false statement, as ${ \ln 120 }$ is not equal to ${ \ln 30 }$. Therefore, x = 300 is not the solution.

By systematically substituting each incorrect option into the original equation, we've demonstrated that none of them satisfy the equation. This reinforces the validity of our solution, x = 75, and provides a comprehensive understanding of the problem-solving process.

Mastering Logarithmic Equations: Key Takeaways

Solving logarithmic equations, like the one we've tackled, is a fundamental skill in mathematics. By understanding and applying the properties of logarithms, we can transform complex equations into simpler forms, leading us to the correct solution. Here are some key takeaways from our exploration:

1. Know Your Properties: A solid grasp of logarithmic properties, such as the power rule, quotient rule, product rule, and the inverse relationship between logarithms and exponentials, is crucial.
2. Simplify Step by Step: Break down the equation into manageable steps, simplifying expressions at each stage. This approach prevents errors and makes the solution process more transparent.
3. Check Your Solution: Always substitute your solution back into the original equation to verify its correctness. This step ensures that you haven't made any algebraic errors along the way.
4. Understand Incorrect Options: Analyzing why incorrect options fail to satisfy the equation deepens your understanding of the problem and the solution process.

By mastering these principles, you'll be well-equipped to tackle a wide range of logarithmic equations and mathematical challenges.

This comprehensive guide has not only provided the solution to the equation but also illuminated the underlying principles and techniques for solving logarithmic equations. As you continue your mathematical journey, remember that practice and a deep understanding of fundamental properties are key to success.