Solving Logarithmic Equation Log₂(6x) - Log₂(√x) = 2 A Step-by-Step Guide

Jul 12, 2025 by Admin 74 views

Unlocking the Solution to the Logarithmic Equation: log₂(6x) - log₂(√x) = 2

Introduction: Delving into Logarithmic Equations

In the realm of mathematics, logarithmic equations hold a significant place, often appearing in various scientific and engineering applications. These equations involve logarithms, which are essentially the inverse operations of exponentiation. To master these equations, a solid understanding of logarithmic properties and algebraic manipulation is crucial. In this article, we will dissect a specific logarithmic equation, $\log _2(6 x)-\log _2(\sqrt{x})=2$ , and explore the step-by-step process to arrive at the correct solution. We will not only solve the equation but also delve into the underlying concepts and potential pitfalls, ensuring a comprehensive understanding of the topic. This journey into the world of logarithms will equip you with the necessary tools to tackle similar problems with confidence.

Understanding Logarithms: The Foundation

Before we jump into solving the equation, let's take a moment to solidify our understanding of logarithms. A logarithm answers the question: "To what power must we raise a base to obtain a certain number?" Mathematically, if $b^y = x$ , then we can write this relationship in logarithmic form as $\log_b(x) = y$ , where 'b' is the base, 'x' is the argument, and 'y' is the exponent. Understanding this fundamental relationship is the key to unraveling logarithmic equations. The base of a logarithm plays a vital role in its behavior. Common bases include 10 (common logarithm) and 'e' (natural logarithm), but in our case, the base is 2. The argument of a logarithm must be a positive number, as logarithms are not defined for non-positive values. This is a crucial constraint that we need to keep in mind when solving logarithmic equations, as it can lead to extraneous solutions. Furthermore, logarithmic functions possess certain properties that are invaluable in simplifying and solving equations. These properties include the product rule, quotient rule, and power rule, which we will utilize in the subsequent sections.

Problem Statement: Decoding the Equation

The equation at hand is $\log _2(6 x)-\log _2(\sqrt{x})=2$ . This equation presents a challenge that requires us to utilize the properties of logarithms to simplify and isolate the variable 'x'. The left-hand side of the equation involves the difference of two logarithms with the same base, which immediately suggests the application of the quotient rule of logarithms. The right-hand side is a constant, which we will eventually need to express in logarithmic form to equate the arguments. A critical aspect to consider when dealing with logarithmic equations is the domain of the logarithmic functions. The arguments of the logarithms, $6x$ and $\sqrt{x}$ , must be positive. This implies that $x > 0$ , which will be an important constraint when we arrive at potential solutions. By carefully dissecting the equation and identifying the key components, we can chart a course towards the solution. The presence of both a linear term ( $6x$ ) and a square root term ( $\sqrt{x}$ ) within the logarithms adds a layer of complexity, but with a systematic approach, we can overcome this hurdle.

Solving the Equation: A Step-by-Step Approach

Now, let's embark on the journey of solving the logarithmic equation $\log _2(6 x)-\log _2(\sqrt{x})=2$ . Our strategy will involve applying logarithmic properties, simplifying the equation, and ultimately isolating 'x'.

Applying the Quotient Rule: The quotient rule of logarithms states that $\log_b(m) - \log_b(n) = \log_b(\frac{m}{n})$ . Applying this rule to our equation, we get:
$\log _2\left(\frac{6 x}{\sqrt{x}}\right)=2$
This step elegantly combines the two logarithms into a single logarithm, making the equation more manageable. The quotient rule is a powerful tool in simplifying logarithmic expressions, and its application here is a crucial first step.
Simplifying the Argument: We can simplify the argument of the logarithm by rationalizing the denominator:
$\frac{6 x}{\sqrt{x}} = \frac{6 x \cdot \sqrt{x}}{\sqrt{x} \cdot \sqrt{x}} = \frac{6 x \sqrt{x}}{x} = 6 \sqrt{x}$
Therefore, our equation now becomes:
$\log _2(6 \sqrt{x})=2$
This simplification is key to isolating 'x'. By eliminating the fraction within the logarithm, we pave the way for further manipulation.
Converting to Exponential Form: To eliminate the logarithm, we convert the equation to its exponential form. Recall that $\log_b(x) = y$ is equivalent to $b^y = x$ . Applying this to our equation, we get:
$2^2 = 6 \sqrt{x}$ $4 = 6 \sqrt{x}$
This step is the heart of solving logarithmic equations. By converting to exponential form, we free 'x' from the confines of the logarithm.
Isolating the Square Root: To isolate the square root term, we divide both sides of the equation by 6:
$\frac{4}{6} = \sqrt{x}$ $\frac{2}{3} = \sqrt{x}$
Isolating the square root is a necessary step before we can square both sides and solve for 'x'.
Squaring Both Sides: To eliminate the square root, we square both sides of the equation:
$\left(\frac{2}{3}\right)^2 = (\sqrt{x})^2$ $\frac{4}{9} = x$
This step directly yields the value of 'x'. Squaring both sides is a common technique for solving equations involving square roots, but it's crucial to remember to check for extraneous solutions.
Checking for Extraneous Solutions: We must check if our solution, $x = \frac{4}{9}$ , satisfies the original equation and the domain restrictions. Recall that the arguments of the logarithms, $6x$ and $\sqrt{x}$ , must be positive. Since $\frac{4}{9} > 0$ , this condition is met. Now, let's substitute $x = \frac{4}{9}$ back into the original equation:
$\log _2\left(6 \cdot \frac{4}{9}\right)-\log _2\left(\sqrt{\frac{4}{9}}\right)=2$ $\log _2\left(\frac{8}{3}\right)-\log _2\left(\frac{2}{3}\right)=2$
Applying the quotient rule again:
$\log _2\left(\frac{8/3}{2/3}\right)=2$ $\log _2(4)=2$ $2 = 2$
The equation holds true, so $x = \frac{4}{9}$ is a valid solution. This crucial step of checking for extraneous solutions ensures the integrity of our answer.

Analyzing the Options: Finding the Correct Answer

Now that we have solved the equation and found the solution $x = \frac{4}{9}$ , let's examine the given options:

A. $x=0$ B. $x=\frac{2}{9}$ C. $x=\frac{4}{9}$ D. $x=\frac{2}{3}$

Clearly, option C, $x = \frac{4}{9}$ , matches our solution. The other options are incorrect. Option A, $x = 0$ , is particularly problematic because it would make the arguments of the logarithms undefined. Options B and D are simply numerical values that do not satisfy the equation.

Common Pitfalls: Avoiding Mistakes

When solving logarithmic equations, there are several common pitfalls that students often encounter. Awareness of these potential errors can significantly improve accuracy.

Forgetting Domain Restrictions: The most common mistake is neglecting the domain restrictions of logarithms. The argument of a logarithm must be positive. Failing to check this can lead to extraneous solutions.
Incorrectly Applying Logarithmic Properties: The properties of logarithms, such as the product rule, quotient rule, and power rule, must be applied correctly. Misapplication of these rules can lead to incorrect simplifications and solutions.
Not Checking for Extraneous Solutions: Squaring both sides of an equation or performing other operations that can introduce extraneous solutions necessitates a check by substituting the potential solutions back into the original equation.
Algebraic Errors: Simple algebraic errors, such as incorrect simplification or manipulation, can derail the entire solution process. Careful attention to detail is crucial.
Misunderstanding the Definition of Logarithms: A solid understanding of the fundamental relationship between logarithms and exponents is essential. Confusion about this relationship can lead to errors in converting between logarithmic and exponential forms.

By being mindful of these pitfalls and practicing diligently, you can significantly enhance your ability to solve logarithmic equations accurately and efficiently.

Conclusion: Mastering Logarithmic Equations

In this article, we have meticulously solved the logarithmic equation $\log _2(6 x)-\log _2(\sqrt{x})=2$ . We began by establishing a solid understanding of logarithms and their properties. We then systematically applied these properties to simplify the equation, isolate the variable, and arrive at the solution $x = \frac{4}{9}$ . Furthermore, we emphasized the importance of checking for extraneous solutions and highlighted common pitfalls to avoid. Mastering logarithmic equations requires a combination of conceptual understanding, algebraic skill, and careful attention to detail. By following a structured approach and practicing regularly, you can confidently tackle these equations and excel in your mathematical endeavors. The journey through this problem has provided valuable insights into the nature of logarithms and their applications. With a firm grasp of these concepts, you are well-equipped to navigate the complexities of logarithmic equations and related mathematical challenges.