Solving Logarithmic Equations A Step-by-Step Guide

Aug 2, 2025 by Omar Yusuf 51 views

$Solving the Logarithmic Equation: $\log_4 x + \log_4(x+6) = 2$$

Hey guys! Let's dive into this logarithmic equation and figure out the potential solutions. We've got $\log_4 x + \log_4(x+6) = 2$ , and we need to find the values of x that make this equation true. It looks a bit intimidating at first, but don't worry, we'll break it down step by step. Our goal is to understand how to manipulate logarithmic equations, apply the properties of logarithms, and ultimately, solve for x. We’ll also need to be mindful of the domain of logarithmic functions, ensuring our solutions are valid.

Understanding Logarithmic Equations

To solve logarithmic equations effectively, it's crucial to understand the basics of logarithms. A logarithm is essentially the inverse operation of exponentiation. The expression $\log_b a = c$ means that $b^c = a$ . In our equation, we have logarithms with base 4, so we're dealing with powers of 4. Logarithmic functions have a restricted domain; specifically, the argument of a logarithm must be positive. This means that in our equation, both x and x + 6 must be greater than zero. This constraint will be important when we check our solutions later.

Applying Logarithmic Properties

The key to solving this equation lies in the properties of logarithms. One of the most useful properties is the product rule, which states that $\log_b m + \log_b n = \log_b (mn)$ . This allows us to combine the two logarithmic terms on the left side of our equation into a single logarithm. By applying this rule, we can simplify the equation and make it easier to work with. Remember, the goal is to isolate the variable x, and using the properties of logarithms is a crucial step in that process. Once we've combined the logarithms, we can then convert the logarithmic equation into an exponential equation, which is usually easier to solve.

Step-by-Step Solution

Let's walk through the solution step by step to make sure we understand every part of the process. This involves manipulating the equation, combining logarithmic terms, converting to exponential form, and solving the resulting quadratic equation. We'll also need to check our solutions against the domain restrictions of the logarithms to ensure they are valid.

1. Combine Logarithms Using the Product Rule

First, we'll use the product rule of logarithms to combine the two terms on the left side of the equation. The equation is $\log_4 x + \log_4(x+6) = 2$ . Applying the product rule, we get:

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

This simplifies the equation by combining the two logarithmic terms into one. This is a crucial step in solving logarithmic equations, as it allows us to eliminate the logarithms and work with a simpler algebraic expression. This step transforms the equation into a form where we can apply the definition of a logarithm to convert it into an exponential equation.

2. Convert to Exponential Form

Next, we'll convert the logarithmic equation to its equivalent exponential form. Remember that $\log_b a = c$ is equivalent to $b^c = a$ . In our case, the base is 4, the exponent is 2, and the argument is x( x + 6). So, we can rewrite the equation as:

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

This step is vital because it removes the logarithm, allowing us to work with a regular algebraic equation. Converting to exponential form is a standard technique for solving logarithmic equations, and it's a step you'll use frequently.

3. Simplify and Rearrange

Now, let's simplify the equation and rearrange it into a standard quadratic form. We have:

$16 = x^2 + 6x$

To get it into the standard quadratic form, we'll subtract 16 from both sides:

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

This step involves basic algebraic manipulation to get the equation into a form that we can easily solve. A quadratic equation in the form $ax^2 + bx + c = 0$ can be solved by factoring, completing the square, or using the quadratic formula.

4. Solve the Quadratic Equation

We now have a quadratic equation, $x^2 + 6x - 16 = 0$ . We can solve this by factoring. We're looking for two numbers that multiply to -16 and add to 6. Those numbers are 8 and -2. So, we can factor the quadratic as:

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

This gives us two potential solutions for x:

$x + 8 = 0$ or $x - 2 = 0$

Solving these, we get:

$x = -8$ or $x = 2$

Factoring is a common method for solving quadratic equations, and it's often the quickest method when the quadratic can be factored easily. The zero-product property, which states that if the product of two factors is zero, then at least one of the factors must be zero, is the key principle behind this step.

5. Check for Extraneous Solutions

It's crucial to check our solutions in the original equation because logarithms are only defined for positive arguments. We need to make sure that x and x + 6 are both greater than zero.

For x = -8:
- $\log_4 (-8)$ is undefined because the argument is negative.
- So, x = -8 is an extraneous solution.
For x = 2:
- $\log_4 (2)$ is defined.
- $\log_4 (2 + 6) = \log_4 (8)$ is defined.
- So, x = 2 is a valid solution.

Checking for extraneous solutions is a vital step in solving logarithmic equations. Logarithmic functions have domain restrictions, and it's possible to obtain solutions that do not satisfy these restrictions. These solutions are called extraneous solutions and must be discarded.

Final Answer

After checking our solutions, we find that only x = 2 is a valid solution. The solution x = -8 is extraneous because it results in taking the logarithm of a negative number, which is undefined. Therefore, the only solution to the equation $\log_4 x + \log_4(x+6) = 2$ is x = 2.

So, the correct answer is:

C. x = 2 and x = -8

Why This Answer Matters

Understanding how to solve logarithmic equations is super important in math and science. These equations pop up in all sorts of places, from calculating pH levels in chemistry to modeling population growth in biology. Being able to confidently tackle these problems means you're building a strong foundation for more advanced topics. It’s not just about getting the right answer; it’s about understanding the process and why it works. By breaking down the problem step by step, you’re developing critical thinking skills that will help you in all areas of life.

Tips for Mastering Logarithmic Equations

Here are a few extra tips to help you become a pro at solving logarithmic equations:

Know Your Properties: Make sure you're comfortable with all the properties of logarithms, like the product rule, quotient rule, and power rule. These are your best friends when it comes to simplifying and solving equations.
Practice Makes Perfect: The more you practice, the better you'll get. Work through a variety of problems, and don't be afraid to make mistakes – that's how you learn!
Check Your Answers: Always, always, always check your solutions in the original equation. This is the best way to catch extraneous solutions and ensure you've got the right answer.
Understand the Domain: Remember that the argument of a logarithm must be positive. Keep this in mind when setting up and solving equations.

By following these tips and practicing regularly, you'll be solving logarithmic equations like a champ in no time!

Conclusion

We've successfully navigated the process of solving the logarithmic equation $\log_4 x + \log_4(x+6) = 2$ . We started by understanding the properties of logarithms, then applied these properties to simplify the equation. We converted the logarithmic equation to exponential form, solved the resulting quadratic equation, and crucially, checked our solutions for extraneous values. This thorough approach ensures we arrive at the correct solution: x = 2. Remember, logarithmic equations might seem tricky, but with a systematic approach and a solid understanding of logarithmic properties, you can solve them confidently. Keep practicing, and you'll master these types of problems in no time. Understanding these concepts thoroughly prepares you for more advanced mathematical topics and real-world applications where logarithmic functions play a crucial role.