Solving Logarithmic Equations A Step By Step Guide To Log4 X + Log4(x+6) = 2

Hey there, math enthusiasts! Today, we're diving deep into the world of logarithms to tackle a fascinating equation: $\log_4 x + \log_4(x+6) = 2$. This problem isn't just about crunching numbers; it's about understanding the fundamental properties of logarithms and how they interact with algebraic equations. Whether you're a student prepping for an exam or simply a lover of mathematical puzzles, this step-by-step guide will break down the solution in a clear, concise, and engaging way. Let's embark on this mathematical journey together!

Understanding Logarithms: The Key to Unlocking the Equation

Before we jump into solving the equation, let's take a moment to refresh our understanding of logarithms. At its heart, a logarithm answers the question: "To what power must we raise a base to get a certain number?" In our equation, the base is 4. So, when we see $\log_4 x$, we're essentially asking, "To what power must we raise 4 to get x?" Grasping this concept is crucial because logarithms are the inverse operation of exponentiation. This inverse relationship is what allows us to simplify and solve logarithmic equations.

Logarithmic Properties: The Foundation of Our Solution

To effectively solve our equation, we need to leverage some key logarithmic properties. These properties act as tools in our mathematical toolkit, allowing us to manipulate and simplify expressions. The most important property for this problem is the product rule of logarithms, which states:

$$\log_b m + \log_b n = \log_b (mn)$$

In simpler terms, the logarithm of the product of two numbers is equal to the sum of the logarithms of those numbers (provided they have the same base). This property is a game-changer because it allows us to combine the two logarithmic terms on the left side of our equation into a single logarithm. Additionally, we'll use the fundamental relationship between logarithms and exponents to eliminate the logarithm altogether. Remember, $\log_b a = c$ is equivalent to $b^c = a$. This conversion is the key to unlocking the variable x from within the logarithmic expression.

Potential Pitfalls: Domains and Extraneous Solutions

Now, a word of caution before we proceed. Logarithmic functions have a restricted domain. This means that not all values can be plugged into a logarithm. Specifically, we can only take the logarithm of positive numbers. Why? Because there's no power to which you can raise a positive base (like 4) to get a negative number or zero. This domain restriction is critical because it means that we need to check our solutions at the end to make sure they're valid. We might find solutions that algebraically satisfy the equation but are not actually solutions because they fall outside the domain of the logarithm. These are called extraneous solutions, and identifying and discarding them is a crucial step in solving logarithmic equations.

Step-by-Step Solution: Cracking the Code

Alright, with our understanding of logarithms and their properties in place, let's tackle the equation head-on. Here's a detailed, step-by-step solution to $\log_4 x + \log_4(x+6) = 2$:

Step 1: Applying the Product Rule

The first thing we're going to do is simplify the left side of the equation using the product rule of logarithms. Remember, this rule allows us to combine two logarithms with the same base that are being added together. Applying the rule, we get:

$$\log_4 x + \log_4(x+6) = \log_4 [x(x+6)]$$

So, our equation now looks like this:

$$\log_4 [x(x+6)] = 2$$

Step 2: Converting to Exponential Form

Next, we'll eliminate the logarithm by converting the equation to its exponential form. Recall that $\log_b a = c$ is equivalent to $b^c = a$. Applying this to our equation, we get:

$$4^2 = x(x+6)$$

This simplifies to:

$$16 = x(x+6)$$

Step 3: Simplifying and Rearranging

Now we have a simple algebraic equation to solve. Let's expand the right side and rearrange the equation into a standard quadratic form:

$$16 = x^2 + 6x$$

Subtracting 16 from both sides, we get:

$$x^2 + 6x - 16 = 0$$

Step 4: Factoring the Quadratic

We now have a quadratic equation in the form $ax^2 + bx + c = 0$. To solve this, we can try factoring. We're looking for two numbers that multiply to -16 and add up to 6. Those numbers are 8 and -2. So, we can factor the quadratic as follows:

$$(x + 8)(x - 2) = 0$$

Step 5: Finding Potential Solutions

To find the potential solutions for x, we set each factor equal to zero:

$$x + 8 = 0 \quad \text{or} \quad x - 2 = 0$$

Solving these equations, we get two potential solutions:

$$x = -8 \quad \text{or} \quad x = 2$$

Step 6: Checking for Extraneous Solutions

This is the crucial step where we check our potential solutions against the domain of the logarithms in the original equation. Remember, we can only take the logarithm of positive numbers. Let's check each solution:

• For x = -8:
  • $\log_4(-8)$ is undefined because we can't take the logarithm of a negative number.
  • $\log_4(-8 + 6) = \log_4(-2)$ is also undefined for the same reason.
  • Therefore, x = -8 is an extraneous solution and must be discarded.
• For x = 2:
  • $\log_4(2)$ is defined (2 is positive).
  • $\log_4(2 + 6) = \log_4(8)$ is also defined (8 is positive).
  • Therefore, x = 2 is a valid solution.

Step 7: The Final Solution

After carefully checking for extraneous solutions, we've arrived at our final answer. The only valid solution to the equation $\log_4 x + \log_4(x+6) = 2$ is:

$$x = 2$$

Analyzing the Answer Choices: Identifying the Correct Option

Now that we've found the solution, let's circle back to the original answer choices and identify the correct one. The options were:

• A. $x=-2$ and $x=-8$
• B. $x=-2$ and $x=8$
• C. $x=2$ and $x=-8$
• D. $x=2$ and $x=8$

Based on our step-by-step solution, we determined that the only valid solution is $x = 2$. Therefore, the correct answer choice is:

• C. $x=2$ and $x=-8$

Even though the answer choice includes $x=-8$, we know this is an extraneous solution, and $x=2$ is the valid answer.

Common Mistakes and How to Avoid Them

When working with logarithmic equations, there are a few common pitfalls that students often encounter. Being aware of these mistakes can help you avoid them and ensure you arrive at the correct solution. Let's explore some of these common errors and discuss how to steer clear of them.

Forgetting the Domain Restriction

One of the most frequent mistakes is forgetting that logarithms are only defined for positive numbers. As we saw in our solution, this means that we need to check our potential solutions to make sure they don't result in taking the logarithm of a negative number or zero. If a solution does, it's an extraneous solution and must be discarded. To avoid this mistake, always remember to check your solutions against the domain of the original logarithmic equation.

Incorrectly Applying Logarithmic Properties

Logarithmic properties are powerful tools, but they need to be applied correctly. A common error is misapplying the product, quotient, or power rules. For example, students might mistakenly think that $\log_b(m + n)$ is equal to $\log_b m + \log_b n$, which is incorrect. The product rule only applies to the logarithm of a product, not a sum. To avoid these errors, take the time to thoroughly understand the logarithmic properties and practice applying them in various situations. Double-check your work to ensure you're using the properties correctly.

Errors in Algebraic Manipulation

Solving logarithmic equations often involves algebraic manipulation, such as expanding expressions, factoring quadratics, or solving linear equations. Mistakes in these steps can lead to incorrect solutions. For instance, a sign error when rearranging terms or an incorrect factorization can throw off the entire solution process. To minimize these errors, be meticulous in your algebraic steps. Write out each step clearly and double-check your work as you go. Practice your algebra skills regularly to build confidence and accuracy.

Not Checking for Extraneous Solutions

Even if you correctly apply logarithmic properties and algebraic techniques, you might still end up with an incorrect answer if you fail to check for extraneous solutions. As we saw in our example, one of the potential solutions we found algebraically was not a valid solution because it fell outside the domain of the logarithm. To avoid this mistake, make it a habit to always check your solutions by plugging them back into the original equation. If a solution results in taking the logarithm of a non-positive number, it's extraneous and should be discarded.

Misunderstanding the Definition of a Logarithm

At its core, solving logarithmic equations requires a solid understanding of what a logarithm actually represents. If you're unclear on the fundamental definition – that $\log_b a = c$ means $b^c = a$ – you might struggle to convert between logarithmic and exponential forms, which is a crucial step in solving many logarithmic equations. To strengthen your understanding, revisit the definition of a logarithm and practice converting logarithmic expressions to exponential form and vice versa. The more comfortable you are with this fundamental concept, the easier it will be to solve logarithmic equations.

Conclusion: Mastering Logarithmic Equations

We've journeyed through the solution of the equation $\log_4 x + \log_4(x+6) = 2$, and along the way, we've reinforced our understanding of logarithms, logarithmic properties, and the importance of checking for extraneous solutions. Remember, logarithmic equations might seem daunting at first, but with a systematic approach and a solid grasp of the underlying principles, you can conquer them with confidence. Keep practicing, and you'll become a master of logarithmic equations in no time!

So, there you have it, folks! We've successfully solved a tricky logarithmic equation, and hopefully, you've gained some valuable insights along the way. Keep those math muscles flexed, and remember, every problem is just a puzzle waiting to be solved. Until next time, happy problem-solving!