Solving Logarithmic Equations Expressing X In Terms Of A

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Introduction

In this comprehensive article, we delve into the intricacies of solving logarithmic equations, specifically focusing on expressing x in terms of a given the equation:

${ \log_a \left( \frac{x}{a^2} \right) = 3 \log_a 2 - \log_a (x - 2a) }$

Logarithmic equations often appear daunting, but with a systematic approach and a solid understanding of logarithmic properties, they can be tackled effectively. This article aims to provide a step-by-step solution, ensuring clarity and comprehension for readers of all levels. We will begin by outlining the fundamental logarithmic properties essential for solving this equation. Subsequently, we will apply these properties to simplify the given equation, isolate x, and express it in terms of a. Furthermore, we will discuss the importance of verifying solutions in the context of logarithmic equations, as extraneous solutions can arise due to the domain restrictions of logarithmic functions. Through detailed explanations and practical examples, this article will equip you with the skills necessary to confidently solve similar logarithmic problems.

Fundamental Logarithmic Properties

Before diving into the solution, it's crucial to understand the core logarithmic properties that will guide our steps. These properties allow us to manipulate logarithmic expressions, combining or separating terms as needed. Here are the key properties we'll utilize:

  1. Product Rule:

    ${ \log_b (MN) = \log_b M + \log_b N }$

    The logarithm of a product is the sum of the logarithms of the individual factors. This property allows us to expand a single logarithmic term into multiple terms, each involving a factor of the original argument. Conversely, it can be used to combine separate logarithmic terms into a single term involving the product of their arguments.

  2. Quotient Rule:

    ${ \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N }$

    The logarithm of a quotient is the difference between the logarithms of the numerator and the denominator. This property is the counterpart to the product rule, allowing us to deal with division within a logarithmic expression. It is particularly useful when the argument of a logarithm is a fraction, as in the given equation where we have log⁑a(xa2){ \log_a \left( \frac{x}{a^2} \right) }.

  3. Power Rule:

    ${ \log_b (M^p) = p \log_b M }$

    The logarithm of a number raised to a power is the product of the power and the logarithm of the number. This property is invaluable for simplifying expressions where the argument of a logarithm involves an exponent. In the given equation, the term 3log⁑a2{ 3 \log_a 2 } can be simplified using this rule, making it easier to combine with other logarithmic terms.

  4. Change of Base Rule:

    ${ \log_b M = \frac{\log_c M}{\log_c b} }$

    While not directly used in solving this particular equation, the change of base rule is a fundamental property that allows us to convert logarithms from one base to another. This is particularly useful when dealing with calculators that may not support logarithms of arbitrary bases. Understanding this rule provides a more complete understanding of logarithmic operations.

  5. Logarithm of Base:

    ${ \log_b b = 1 }$

    The logarithm of the base itself is always 1. This property stems directly from the definition of a logarithm and is often used in simplifying expressions where the argument is equal to the base.

  6. Logarithm of 1:

    ${ \log_b 1 = 0 }$

    The logarithm of 1 to any base is always 0. This is because any number raised to the power of 0 equals 1. This property is useful when simplifying equations where a logarithmic term is equal to 0.

  7. Inverse Property:

    ${ b^{\log_b M} = M }$

    Exponentiating a logarithm with the same base as the exponent cancels out, leaving the original argument. This property is a powerful tool for solving logarithmic equations, as it allows us to eliminate logarithms and obtain an algebraic equation.

Understanding and mastering these logarithmic properties is essential for solving various logarithmic equations. In the following sections, we will apply these properties step-by-step to solve the given equation and express x in terms of a. Each step will be explained in detail, ensuring a clear understanding of the process. By the end of this article, you will have a solid grasp of how to approach and solve similar logarithmic problems effectively.

Step-by-Step Solution

Now, let's apply these properties to the given equation:

${ \log_a \left( \frac{x}{a^2} \right) = 3 \log_a 2 - \log_a (x - 2a) }$

Our goal is to isolate x and express it in terms of a. We'll proceed systematically, utilizing the logarithmic properties we've discussed.

Step 1: Applying the Quotient Rule

We begin by applying the quotient rule to the left-hand side of the equation:

${ \log_a \left( \frac{x}{a^2} \right) = \log_a x - \log_a a^2 }$

This step separates the logarithm of the quotient into the difference of the logarithms of the numerator and the denominator. It allows us to simplify the expression further by dealing with each term individually.

Step 2: Applying the Power Rule

Next, we apply the power rule to the second term on the left-hand side:

${ \log_a a^2 = 2 \log_a a }$

And we know that log⁑aa=1{ \log_a a = 1 }, so:

${ 2 \log_a a = 2 \cdot 1 = 2 }$

This simplification makes the equation more manageable by removing the exponent within the logarithm.

Step 3: Applying the Power Rule on the Right-Hand Side

Now, let's focus on the right-hand side of the original equation. We apply the power rule to the term 3log⁑a2{ 3 \log_a 2 }:

${ 3 \log_a 2 = \log_a 2^3 = \log_a 8 }$

This step consolidates the coefficient into the argument of the logarithm, preparing us to combine the logarithmic terms.

Step 4: Rewriting the Equation

Substituting these simplifications back into the original equation, we get:

${ \log_a x - 2 = \log_a 8 - \log_a (x - 2a) }$

This rewritten equation is now in a more concise form, allowing us to proceed with combining the logarithmic terms.

Step 5: Combining Logarithmic Terms

To combine the logarithmic terms, we move the log⁑a(xβˆ’2a){ \log_a (x - 2a) } term to the left-hand side:

${ \log_a x + \log_a (x - 2a) = \log_a 8 + 2 }$

Now, we apply the product rule to combine the logarithms on the left-hand side:

${ \log_a [x(x - 2a)] = \log_a 8 + 2 }$

Step 6: Eliminating the Logarithm

To eliminate the logarithm, we need to rewrite the constant 2 as a logarithm with base a. Recall that log⁑aa=1{ \log_a a = 1 }, so 2=2log⁑aa=log⁑aa2{ 2 = 2 \log_a a = \log_a a^2 }. Therefore, we can rewrite the equation as:

${ \log_a [x(x - 2a)] = \log_a 8 + \log_a a^2 }$

Applying the product rule to the right-hand side:

${ \log_a [x(x - 2a)] = \log_a (8a^2) }$

Since the logarithms are equal, their arguments must be equal as well:

${ x(x - 2a) = 8a^2 }$

Step 7: Solving the Quadratic Equation

Expanding the left-hand side, we get a quadratic equation:

${ x^2 - 2ax = 8a^2 }$

Rearranging the terms to set the equation to zero:

${ x^2 - 2ax - 8a^2 = 0 }$

Now, we can factor the quadratic equation:

${ (x - 4a)(x + 2a) = 0 }$

This gives us two potential solutions for x:

${ x = 4a }$ or ${ x = -2a }$

Step 8: Verifying the Solutions

It's crucial to verify these solutions in the original equation because logarithmic functions have domain restrictions. The argument of a logarithm must be positive.

Case 1: x=4a{ x = 4a }

Substituting x=4a{ x = 4a } into the original equation, we need to ensure that both { rac{x}{a^2} } and xβˆ’2a{ x - 2a } are positive.

${ rac{4a}{a^2} = \frac{4}{a} > 0 }$

This implies that a>0{ a > 0 }.

${ 4a - 2a = 2a > 0 }$

This also implies that a>0{ a > 0 }. Thus, x=4a{ x = 4a } is a valid solution when a>0{ a > 0 }.

Case 2: x=βˆ’2a{ x = -2a }

Substituting x=βˆ’2a{ x = -2a } into the original equation:

${ \frac{-2a}{a^2} = \frac{-2}{a} }$

For this to be positive, a<0{ a < 0 }.

${ -2a - 2a = -4a }$

For βˆ’4a>0{ -4a > 0 }, we also need a<0{ a < 0 }. However, the term log⁑a(xa2){ \log_a \left( \frac{x}{a^2} \right) } requires a>0{ a > 0 } and aβ‰ 1{ a \neq 1 } for the logarithm to be defined. Therefore, x=βˆ’2a{ x = -2a } is an extraneous solution.

Final Answer

Therefore, the only valid solution is:

${ x = 4a }$

This solution is valid provided that a>0{ a > 0 }.

Importance of Checking for Extraneous Solutions

In the realm of logarithmic equations, the process of verifying solutions holds paramount importance due to the inherent domain restrictions of logarithmic functions. The argument of a logarithm, the expression inside the logarithmic function, must always be positive. This restriction arises from the very definition of logarithms as the inverse of exponential functions. Since exponential functions yield only positive values, their inverse, the logarithm, can only accept positive inputs.

When solving logarithmic equations, we often employ algebraic manipulations, such as combining logarithmic terms or exponentiating both sides of the equation. While these manipulations are essential for isolating the variable, they can sometimes introduce extraneous solutionsβ€”values that satisfy the transformed equation but not the original one. These extraneous solutions typically arise when the algebraic manipulations lead to expressions that are defined for a broader range of values than the original logarithmic expressions.

Consider the initial equation we tackled in this article:

${ \log_a \left( \frac{x}{a^2} \right) = 3 \log_a 2 - \log_a (x - 2a) }$

In the process of solving, we arrived at two potential solutions: x=4a{ x = 4a } and x=βˆ’2a{ x = -2a }. However, upon verification, we found that x=βˆ’2a{ x = -2a } was an extraneous solution. This occurred because substituting x=βˆ’2a{ x = -2a } into the original equation resulted in taking the logarithm of a non-positive number, violating the domain restriction of the logarithmic function. Specifically, the term xβˆ’2a{ x - 2a } became negative when x=βˆ’2a{ x = -2a }, rendering the solution invalid.

The act of checking for extraneous solutions is not merely a formality; it is an integral part of the problem-solving process. It ensures that the solutions we obtain are not only algebraically correct but also meaningful in the context of the original equation. This step is particularly crucial in logarithmic equations, where the domain restrictions are non-intuitive and easily overlooked.

To effectively check for extraneous solutions, each potential solution must be substituted back into the original equation. We must then verify that all logarithmic terms are defined, meaning that their arguments are strictly positive. If any logarithmic term has a non-positive argument, the solution is extraneous and must be discarded. This verification step may seem tedious, but it is a necessary safeguard against incorrect answers.

In addition to the domain restriction of logarithmic functions, extraneous solutions can also arise from other sources, such as squaring both sides of an equation or multiplying by an expression that could be zero. In such cases, the same principle applies: each potential solution must be substituted back into the original equation to ensure its validity. By consistently checking for extraneous solutions, we can maintain the integrity of our mathematical reasoning and arrive at accurate solutions to logarithmic and other types of equations.

Conclusion

In this article, we have provided a detailed, step-by-step solution to the logarithmic equation:

${ \log_a \left( \frac{x}{a^2} \right) = 3 \log_a 2 - \log_a (x - 2a) }$

We successfully expressed x in terms of a, arriving at the solution x=4a{ x = 4a }, which is valid for a>0{ a > 0 }. Along the way, we emphasized the crucial logarithmic properties and their application in solving such equations. We also highlighted the importance of verifying solutions to avoid extraneous results, which is a common pitfall in logarithmic equations.

By understanding and applying the logarithmic properties, one can systematically solve complex logarithmic equations. Remember to always check the domain restrictions and verify your solutions to ensure accuracy. With practice, you can master these techniques and confidently tackle a wide range of logarithmic problems. The ability to solve logarithmic equations is not only a valuable skill in mathematics but also has applications in various fields such as physics, engineering, and computer science. The principles and methods discussed in this article provide a solid foundation for further exploration and application of logarithmic functions in diverse contexts.