Solving The Logarithmic Equation Log2(6x) - Log2(√x) = 2

Let's delve into the world of logarithmic equations and explore how to solve them effectively. In this article, we'll tackle the equation $\log _2(6 x)-\log _2(\sqrt{x})=2$ step by step, providing a comprehensive explanation of each stage. We will start by simplifying the equation using logarithmic properties, then convert it into an exponential form, and finally solve for x. We'll also discuss the importance of verifying solutions in the original equation to avoid extraneous solutions. Understanding these methods is crucial not only for solving specific problems but also for grasping the underlying principles of logarithms, which are fundamental in many areas of mathematics and science. Logarithmic equations, in particular, frequently arise in the context of exponential growth and decay models, which are used extensively in fields such as physics, engineering, biology, and economics. Therefore, a strong grasp of how to solve them is essential for anyone working with these types of models. Mastering the techniques presented here will empower you to tackle a wide array of logarithmic equations with confidence. Understanding logarithmic equations requires a firm grasp of the basic properties of logarithms. This includes the product rule, quotient rule, and power rule. Let's begin by understanding the equation we aim to solve.

Problem Statement: $\log _2(6 x)-\log _2(\sqrt{x})=2$

Our main goal is to determine the true solution to the logarithmic equation $\log _2(6 x)-\log _2(\sqrt{x})=2$. The problem presents us with a logarithmic equation, and our task is to isolate the variable x and find its value. The equation involves logarithms with base 2, and we will use the properties of logarithms to simplify the equation and solve for x. We also need to consider any restrictions on the domain of the logarithmic functions involved, since the argument of a logarithm must be positive. This means that both $6x$ and $\sqrt{x}$ must be greater than zero. The given options are:

A. $x=0$

B. $x=\frac{2}{9}$

C. $x=\frac{4}{9}$

D. $x=\frac{2}{3}$

We will go through the necessary steps to solve this logarithmic equation. Our approach will involve simplifying the equation using logarithmic properties, converting it into an exponential equation, and then solving for x. After finding a potential solution, we will also need to check that it satisfies the original equation, as sometimes solutions obtained through algebraic manipulation may not be valid due to the domain restrictions of logarithms. This is a crucial step in solving logarithmic equations. Specifically, we will start by applying the quotient rule of logarithms to combine the two logarithmic terms on the left side of the equation. Then, we will convert the logarithmic equation into an exponential equation. Finally, we will solve the resulting equation for x and check whether the solution is valid.

Step 1: Apply Logarithmic Properties

To begin solving the logarithmic equation, we'll use the quotient rule of logarithms, which states that $\log_b(m) - \log_b(n) = \log_b(\frac{m}{n})$. Applying this rule to our equation, we have:

$\log _2(6 x)-\log _2(\sqrt{x}) = \log_2(\frac{6x}{\sqrt{x}})$

Now, we simplify the fraction inside the logarithm. Recall that $\sqrt{x} = x^{\frac{1}{2}}$. Thus, we can rewrite the expression as:

$\frac{6x}{\sqrt{x}} = \frac{6x}{x^{\frac{1}{2}}}$

Using the properties of exponents, we subtract the exponents when dividing terms with the same base:

$\frac{6x}{x^{\frac{1}{2}}} = 6x^{1-\frac{1}{2}} = 6x^{\frac{1}{2}} = 6\sqrt{x}$

So, our equation now becomes:

$\log_2(6\sqrt{x}) = 2$

This simplified logarithmic equation is easier to work with. The quotient rule of logarithms is a fundamental property that allows us to combine logarithmic terms with the same base that are being subtracted. By applying this rule, we have transformed the original equation into a more manageable form, where we have a single logarithmic term on one side of the equation. The simplification process also involved the use of exponent rules, which are closely related to logarithmic properties. Specifically, we used the rule that $x^a / x^b = x^{a-b}$ to simplify the expression inside the logarithm. This step is crucial because it reduces the complexity of the equation and prepares it for the next step, which is to convert it into an exponential equation.

Step 2: Convert to Exponential Form

Next, we need to convert the logarithmic equation $\log_2(6\sqrt{x}) = 2$ into its equivalent exponential form. Recall that the logarithmic equation $\log_b(a) = c$ is equivalent to the exponential equation $b^c = a$. Applying this to our equation, we get:

$2^2 = 6\sqrt{x}$

Simplifying the left side, we have:

$4 = 6\sqrt{x}$

Now we have a simple equation involving a square root, which we can solve by isolating the square root term and then squaring both sides. Converting the logarithmic equation to exponential form is a critical step because it eliminates the logarithm and allows us to work with a more familiar algebraic equation. The exponential form directly expresses the relationship between the base, the exponent, and the result of the logarithmic operation. In this case, the base of the logarithm is 2, the exponent is 2, and the result is $6\sqrt{x}$. The conversion is based on the fundamental definition of logarithms, which states that the logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. By applying this definition, we have successfully transformed the logarithmic equation into an algebraic equation involving a square root.

Step 3: Solve for x

Now, let's solve for x in the equation $4 = 6\sqrt{x}$. First, divide both sides by 6:

$\frac{4}{6} = \sqrt{x}$

Simplify the fraction:

$\frac{2}{3} = \sqrt{x}$

To eliminate the square root, square both sides of the equation:

$\left(\frac{2}{3}\right)^2 = (\sqrt{x})^2$

$\frac{4}{9} = x$

So, we have found a potential solution: $x = \frac{4}{9}$. However, it's crucial to check this solution in the original equation to ensure it's valid. Solving for x involves isolating the variable and applying inverse operations to both sides of the equation. In this case, we first divided both sides by 6 to isolate the square root term. Then, we squared both sides to eliminate the square root and obtain a value for x. This process is a standard algebraic technique for solving equations involving radicals. However, it's essential to remember that squaring both sides of an equation can sometimes introduce extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. This is why it is so important to check our solution in the original equation.

Step 4: Check the Solution

We must check if $x = \frac{4}{9}$ is a valid solution by substituting it back into the original logarithmic equation: $\log _2(6 x)-\log _2(\sqrt{x})=2$.

Substitute $x = \frac{4}{9}$:

$\log _2\left(6 \cdot \frac{4}{9}\right)-\log _2\left(\sqrt{\frac{4}{9}}\right) = 2$

Simplify inside the logarithms:

$\log _2\left(\frac{24}{9}\right)-\log _2\left(\frac{2}{3}\right) = 2$

$\log _2\left(\frac{8}{3}\right)-\log _2\left(\frac{2}{3}\right) = 2$

Use the quotient rule of logarithms again:

$\log _2\left(\frac{\frac{8}{3}}{\frac{2}{3}}\right) = 2$

$\log _2\left(\frac{8}{3} \cdot \frac{3}{2}\right) = 2$

$\log _2(4) = 2$

Since $2^2 = 4$, the equation holds true. Therefore, $x = \frac{4}{9}$ is a valid solution. Checking the solution is an indispensable step in solving logarithmic equations. Logarithmic functions have domain restrictions, meaning that the argument of a logarithm must be positive. When we perform algebraic manipulations, such as squaring both sides of an equation, we can sometimes introduce solutions that do not satisfy the original equation's domain restrictions. These solutions are called extraneous solutions. To avoid extraneous solutions, it's crucial to substitute the potential solutions back into the original equation and verify that they make the equation true. In this case, we substituted $x = \frac{4}{9}$ back into the original equation and found that it did indeed satisfy the equation. This confirms that $x = \frac{4}{9}$ is a valid solution.

Final Answer

Therefore, the true solution to the logarithmic equation $\log _2(6 x)-\log _2(\sqrt{x})=2$ is $x = \frac{4}{9}$. So the correct answer is:

C. $x=\frac{4}{9}$

We have successfully solved the logarithmic equation by applying logarithmic properties, converting it into exponential form, solving for x, and verifying the solution. This step-by-step approach ensures that we arrive at the correct answer and avoid any extraneous solutions. The process of solving logarithmic equations involves a combination of algebraic manipulation and an understanding of the fundamental properties of logarithms. By mastering these techniques, you can confidently solve a wide range of logarithmic equations. The key steps include simplifying the equation using logarithmic properties, converting it into exponential form, solving the resulting algebraic equation, and, most importantly, checking the solution in the original equation to ensure its validity. This comprehensive approach not only leads to the correct answer but also deepens your understanding of logarithms and their applications.