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

Hey guys! Today, we're diving into the fascinating world of logarithms and tackling a logarithmic equation step-by-step. Logarithmic equations can seem tricky at first, but with a clear understanding of the rules and properties, they become much more manageable. We'll not only solve the given equation but also explore the underlying concepts to empower you with the knowledge to conquer any logarithmic challenge. So, let's buckle up and embark on this mathematical journey together!

Understanding Logarithms

Before we jump into solving the specific equation, let's take a moment to refresh our understanding of logarithms. Logarithms are essentially the inverse operation of exponentiation. In simpler terms, if we have an exponential equation like ${2^3 = 8}$, the corresponding logarithmic equation would be ${\log_2 8 = 3}$. This reads as "the logarithm base 2 of 8 is 3." The logarithm answers the question: "To what power must we raise the base (2 in this case) to get the argument (8)?"

The general form of a logarithmic equation is ${\log_b a = c}$, where:

• ${b}$ is the base (must be positive and not equal to 1).
• ${a}$ is the argument (must be positive).
• ${c}$ is the exponent.

Understanding this relationship between exponentiation and logarithms is crucial for solving logarithmic equations. Think of logarithms as a way to "undo" exponentiation and vice versa. This inverse relationship allows us to manipulate equations and isolate the variable we're trying to solve for.

Furthermore, several key properties of logarithms are essential tools in our problem-solving arsenal. These properties allow us to simplify complex logarithmic expressions and transform equations into more manageable forms. Let's briefly review some of the most important ones:

1. Product Rule: ${\log_b (mn) = \log_b m + \log_b n}$ - The logarithm of a product is equal to the sum of the logarithms of the individual factors.
2. Quotient Rule: ${\log_b \frac{m}{n} = \log_b m - \log_b n}$ - The logarithm of a quotient is equal to the difference of the logarithms of the numerator and denominator.
3. Power Rule: ${\log_b (m^p) = p \log_b m}$ - The logarithm of a number raised to a power is equal to the power times the logarithm of the number.
4. Change of Base Rule: ${\log_a b = \frac{\log_c b}{\log_c a}}$ - This rule allows us to change the base of a logarithm, which is particularly useful when dealing with calculators that only have logarithms base 10 or base e (natural logarithm).

These properties, especially the product, quotient, and power rules, will be instrumental in simplifying our equation and isolating the variable x. By applying these rules strategically, we can break down complex logarithmic expressions into simpler terms, making the equation easier to solve. Remember, the key is to identify which rule applies best to the specific situation and use it to your advantage. Mastering these properties is like having a Swiss Army knife for logarithmic problems – you'll be equipped to tackle a wide range of challenges!

Problem Restatement: Unpacking the Logarithmic Equation

Okay, let's get down to business! Our mission today is to find the true solution to the following logarithmic equation:

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

Before we start crunching numbers and applying logarithmic properties, it's a good idea to take a moment to really understand what this equation is telling us. We have two logarithmic terms on the left side, both with a base of 2, and they are being subtracted. This subtraction is a major clue, hinting at one of our key logarithmic properties. On the right side, we have a constant value, 2. Our goal is to isolate x, but it's currently trapped inside the logarithmic functions.

To successfully solve this equation, we'll need to carefully unpack it, step-by-step, using the properties of logarithms. We'll start by simplifying the left side of the equation, combining the two logarithmic terms into a single one. This will make the equation much easier to handle. Then, we'll convert the logarithmic equation into its equivalent exponential form. This is a crucial step because it will free x from the clutches of the logarithm. Finally, we'll solve the resulting algebraic equation for x. But we're not done yet! It's super important to check our solution to make sure it's valid. Remember, logarithms are only defined for positive arguments, so we need to ensure that our solution doesn't lead to taking the logarithm of a negative number or zero.

Thinking strategically about the problem before diving into the calculations is a hallmark of a skilled problem-solver. By identifying the key elements of the equation and planning our approach, we can avoid common pitfalls and arrive at the correct solution efficiently. So, let's keep this strategic mindset as we move forward and tackle this logarithmic challenge!

Step-by-Step Solution: Conquering the Logarithmic Challenge

Now, let's roll up our sleeves and get to the heart of the problem – solving the equation ${\log _2(6 x)-\log _2(\sqrt{x})=2}$. We'll break down the solution into a series of clear, manageable steps. By following this methodical approach, we can ensure accuracy and avoid getting lost in the details. So, let's get started!

Step 1: Apply the Quotient Rule of Logarithms

The first thing that likely jumps out is the subtraction between the two logarithmic terms on the left side. This is a clear signal to use the quotient rule of logarithms. Remember, the quotient rule states that ${\log_b \frac{m}{n} = \log_b m - \log_b n}$. Applying this rule to our equation, we can combine the two logarithmic terms into a single logarithm:

${\log _2(6 x)-\log _2(\sqrt{x}) = \log_2 \left(\frac{6x}{\sqrt{x}}\right) = 2}$

This step significantly simplifies the equation. We've gone from having two logarithmic terms to just one, making it much easier to work with. The quotient rule is a powerful tool for simplifying logarithmic expressions, and it's often the first step in solving logarithmic equations.

Step 2: Simplify the Argument of the Logarithm

Now, let's focus on the argument of the logarithm, which is the expression inside the parentheses: ${\frac{6x}{\sqrt{x}}}$. We can simplify this expression by using the properties of exponents. Remember that ${\sqrt{x}}$ is the same as ${x^{\frac{1}{2}}}$. So, we have:

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

To divide terms with the same base, we subtract the exponents:

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

Our equation now looks even simpler:

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

By simplifying the argument of the logarithm, we've made the equation more manageable and brought us closer to isolating x. This step highlights the importance of being comfortable with both logarithmic and exponential properties – they often work hand-in-hand when solving these types of equations.

Step 3: Convert to Exponential Form

Now comes the crucial step of converting the logarithmic equation into its equivalent exponential form. Remember, the logarithmic equation ${\log_b a = c}$ is equivalent to the exponential equation ${b^c = a}$. Applying this to our equation ${\log_2 (6\sqrt{x}) = 2}$, we get:

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

This conversion is the key to unlocking x from the logarithm. We've transformed the equation from a logarithmic form to an algebraic form, which we can now solve using standard algebraic techniques. This step demonstrates the fundamental relationship between logarithms and exponents and how we can leverage this relationship to solve equations.

Step 4: Solve for x

Let's simplify the equation and isolate x. We have:

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

Divide both sides by 6:

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

Simplify the fraction:

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

To get rid of the square root, we square both sides of the equation:

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

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

So, we've found a potential solution: x = ${\frac{4}{9}}$. But we're not quite done yet! We need to check if this solution is valid.

Step 5: Check the Solution

It's absolutely crucial to check our solution in the original equation to make sure it doesn't lead to taking the logarithm of a negative number or zero. Remember, the argument of a logarithm must be positive. Let's plug x = ${\frac{4}{9}}$ back into the original equation:

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

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

Simplify the expressions inside the logarithms:

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

Now, let's apply the quotient rule in reverse to combine the logarithms:

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

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

${\log _2(4)=2}$

Since ${2^2 = 4}$, the equation holds true! Therefore, our solution x = ${\frac{4}{9}}$ is valid.

By following these five steps – applying the quotient rule, simplifying the argument, converting to exponential form, solving for x, and checking the solution – we've successfully navigated the logarithmic equation and found the correct answer. This step-by-step approach is a valuable strategy for tackling any mathematical problem, ensuring clarity, accuracy, and confidence in your solution.

Identifying the Correct Option: The Final Step

Awesome! We've done the hard work and found that the true solution to the logarithmic equation is x = ${\frac{4}{9}}$. Now, let's connect this solution to the given options and pinpoint the correct answer. This final step ensures that we're not just solving the problem but also answering the specific question asked.

Looking back at the options provided:

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

It's clear that our solution, x = ${\frac{4}{9}}$, corresponds to option C. So, we've successfully identified the correct answer!

This final step is a reminder that problem-solving is not just about performing calculations; it's also about understanding the context of the problem and presenting the solution in the required format. In this case, we needed to match our calculated solution to the given options. This attention to detail is crucial for success in mathematics and beyond.

Conclusion: Mastering Logarithmic Equations

Alright, guys! We've reached the end of our logarithmic journey, and what a journey it has been! We've successfully navigated the equation ${\log _2(6 x)-\log _2(\sqrt{x})=2}$ and found the true solution, x = ${\frac{4}{9}}$. But more importantly, we've delved into the core concepts of logarithms, explored their properties, and developed a systematic approach to solving logarithmic equations. This is what true mathematical mastery is all about – not just getting the right answer, but understanding why it's the right answer.

We started by revisiting the fundamental definition of logarithms and their inverse relationship with exponents. We then reviewed the crucial properties of logarithms, particularly the product, quotient, and power rules, which are the building blocks for simplifying logarithmic expressions. With these tools in hand, we tackled the given equation head-on.

Our step-by-step solution demonstrated a strategic approach to problem-solving. We began by applying the quotient rule to combine the logarithmic terms, simplifying the equation. We then simplified the argument of the logarithm using exponent rules. The key step was converting the logarithmic equation into its equivalent exponential form, which allowed us to isolate x. After solving for x, we emphasized the importance of checking the solution to ensure its validity. Finally, we matched our solution to the given options, solidifying our understanding of the problem.

By breaking down the problem into manageable steps, we transformed a seemingly complex equation into a series of simpler tasks. This approach is not only effective for logarithmic equations but also applicable to a wide range of mathematical problems. Remember, the key is to understand the underlying concepts, apply the appropriate tools and techniques, and always check your work.

So, the next time you encounter a logarithmic equation, don't shy away! Embrace the challenge, apply the knowledge you've gained, and confidently conquer it. You've got this!

Remember, practice makes perfect. The more you work with logarithmic equations, the more comfortable and confident you'll become. Keep exploring, keep learning, and keep pushing your mathematical boundaries. The world of logarithms is vast and fascinating, and there's always more to discover. Happy problem-solving!