Solving Logarithmic Equations Potential Solutions For Log4x + Log4(x+6) = 2

by Henrik Larsen 76 views

Hey everyone! Today, we're diving into the fascinating world of logarithms to solve a tricky equation. Logarithmic equations might seem daunting at first, but with a systematic approach, they can be cracked! We're tackling the equation $\log _4 x + \log _4(x+6) = 2$ and exploring the potential solutions. So, buckle up and let's get started!

Understanding Logarithms: The Key to Unlocking the Equation

Before we jump into the solution, let's refresh our understanding of logarithms. Logarithms are essentially the inverse of exponential functions. Think of it this way: if we have an exponential equation like $4^2 = 16$, the logarithm asks the question, "To what power must we raise 4 to get 16?" The answer, of course, is 2. We write this logarithmically as $\log _4 16 = 2$. The small 4 in the subscript is called the base of the logarithm.

In our equation, we have logarithms with a base of 4. The expression $\log _4 x$ means "to what power must we raise 4 to get x?" Similarly, $\log _4(x+6)$ asks, "to what power must we raise 4 to get x+6?" Understanding this fundamental concept is crucial for solving logarithmic equations.

Now, let's talk about some essential properties of logarithms that we'll use to solve the equation. One of the most important properties is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of the individual factors. Mathematically, this is expressed as: $\log _b (mn) = \log _b m + \log _b n$, where b is the base of the logarithm, and m and n are positive numbers. This property will be instrumental in simplifying our equation.

Another key concept to remember is the definition of a logarithm in exponential form. The equation $\log _b a = c$ is equivalent to the exponential equation $b^c = a$. This conversion will be vital in eliminating the logarithms and solving for x. Also, it's crucial to remember that the argument of a logarithm (the expression inside the logarithm) must be positive. This means that in our equation, both x and x+6 must be greater than zero. We'll need to check our solutions later to make sure they satisfy this condition.

Logarithmic equations, like the one we're tackling, often appear in various fields, including mathematics, physics, engineering, and computer science. They're used to model a wide range of phenomena, from exponential growth and decay to the Richter scale for measuring earthquakes. So, mastering the art of solving these equations is a valuable skill.

Solving the Equation: A Step-by-Step Approach

Okay, guys, let's get down to business and solve the equation $\log _4 x + \log _4(x+6) = 2$. We'll break it down step-by-step to make it super clear.

Step 1: Apply the Product Rule of Logarithms

Remember that product rule we talked about? It's time to put it into action! We have the sum of two logarithms with the same base, so we can combine them into a single logarithm using the product rule:

log4x+log4(x+6)=log4[x(x+6)]\\log _4 x + \\log _4(x+6) = \\log _4 [x(x+6)]

So, our equation now looks like this:

log4[x(x+6)]=2\\log _4 [x(x+6)] = 2

Step 2: Convert to Exponential Form

Now, let's get rid of those logarithms altogether! We'll use the definition of a logarithm to rewrite the equation in exponential form. Recall that $\log _b a = c$ is equivalent to $b^c = a$. Applying this to our equation, we get:

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

This simplifies to:

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

Step 3: Simplify and Rearrange into a Quadratic Equation

Let's expand the right side of the equation and rearrange it into a familiar quadratic form:

16=x2+6x16 = x^2 + 6x

Subtracting 16 from both sides, we get:

x2+6x−16=0x^2 + 6x - 16 = 0

Step 4: Solve the Quadratic Equation

We've got a quadratic equation! There are a few ways to solve these. We can try factoring, using the quadratic formula, or even completing the square. In this case, factoring seems like the easiest route. We need to find two numbers that multiply to -16 and add up to 6. Those numbers are 8 and -2. So, we can factor the quadratic as:

(x+8)(x−2)=0(x + 8)(x - 2) = 0

This gives us two potential solutions:

x+8=0Rightarrowx=−8x + 8 = 0 \\Rightarrow x = -8

x−2=0Rightarrowx=2x - 2 = 0 \\Rightarrow x = 2

Step 5: Check for Extraneous Solutions

This is the most important step! Remember that the argument of a logarithm must be positive. We need to plug our potential solutions back into the original equation and see if they work.

Let's start with x = -8:

log4(−8)+log4(−8+6)=log4(−8)+log4(−2)\\log _4 (-8) + \\log _4(-8 + 6) = \\log _4 (-8) + \\log _4 (-2)

Uh oh! We're taking the logarithm of negative numbers. This is a big no-no! So, x = -8 is an extraneous solution – it doesn't actually work in the original equation.

Now, let's check x = 2:

log4(2)+log4(2+6)=log4(2)+log4(8)\\log _4 (2) + \\log _4(2 + 6) = \\log _4 (2) + \\log _4 (8)

Both 2 and 8 are positive, so this solution is potentially valid. Let's see if it satisfies the equation:

log4(2)+log4(8)=log4(2cdot8)=log4(16)=2\\log _4 (2) + \\log _4 (8) = \\log _4 (2 \\cdot 8) = \\log _4 (16) = 2

It works! So, x = 2 is a valid solution.

The Potential Solutions: A Final Answer

After carefully solving the equation and checking for extraneous solutions, we've found that the only valid solution is x = 2.

Therefore, the correct answer is:

  • C. x = 2 and x = -8 (But remember, x = -8 is extraneous!)

It's essential to understand why we include the extraneous solution in the answer choices. It highlights the importance of checking your work and understanding the domain restrictions of logarithmic functions. Always remember to plug your potential solutions back into the original equation to ensure they are valid.

Key Takeaways for Mastering Logarithmic Equations

Before we wrap up, let's recap some key takeaways that will help you conquer logarithmic equations:

  • Understand the Definition of Logarithms: Know what logarithms represent and how they relate to exponential functions. This is the foundation for solving any logarithmic equation.
  • Master Logarithmic Properties: Learn and apply the properties of logarithms, such as the product rule, quotient rule, and power rule. These properties are your tools for simplifying and manipulating logarithmic expressions.
  • Convert to Exponential Form: Use the definition of a logarithm to convert logarithmic equations into exponential equations. This often makes the equation easier to solve.
  • Solve the Resulting Equation: After eliminating the logarithms, you'll usually end up with an algebraic equation (linear, quadratic, etc.). Use your algebraic skills to solve for the variable.
  • Check for Extraneous Solutions: This is crucial! Always plug your potential solutions back into the original logarithmic equation to make sure they are valid. Remember that the argument of a logarithm must be positive.

Solving logarithmic equations might seem tricky at first, but with practice and a solid understanding of the concepts, you'll be solving them like a pro in no time! Keep practicing, and don't hesitate to review the fundamentals whenever you need to. You've got this!

If you found this guide helpful, please feel free to share it with your friends and classmates. And if you have any questions or want to explore more mathematical challenges, let me know in the comments below. Happy problem-solving!