Solving Logarithmic Equations Log Base 2 (x-5) = 3 - Log Base 2 (x-7)

Hey there, math enthusiasts! Today, we're diving into the fascinating world of logarithmic equations. Logarithmic equations can seem a bit intimidating at first, but with a systematic approach, they become quite manageable. We're going to tackle the equation $\log _2(x-5)=3-\log _2(x-7)$ step by step, ensuring you understand not just the how, but also the why behind each move. So, grab your thinking caps, and let's get started!

1. Understanding Logarithms: The Basics

Before we jump into solving our specific equation, let's quickly recap what logarithms are all about. At its heart, a logarithm answers the question: "To what power must we raise the base to get a certain number?" In the expression $\log_b(a) = c$, b is the base, a is the argument (the number we want to get), and c is the exponent (the power we need to raise b to). Put simply, $b^c = a$. This understanding is crucial because logarithms are essentially the inverse operation of exponentiation.

For instance, $\log_2(8) = 3$ because $2^3 = 8$. The base here is 2, the argument is 8, and the exponent is 3. This fundamental relationship between logarithms and exponents is the key to unlocking the secrets of logarithmic equations. When dealing with logarithmic equations, remember that the argument of a logarithm must always be positive. This is because you can't raise a positive base to any power and get a zero or a negative number. This constraint will be particularly important when we check for extraneous solutions later on.

In our equation, we have logarithms with base 2. This means we're looking for powers of 2. The properties of logarithms, which we'll use shortly, allow us to manipulate these expressions and ultimately isolate the variable x. Think of logarithms as tools that help us unravel exponential relationships, making complex equations solvable. Remember, the goal is to use the properties of logarithms to simplify the equation and transform it into a form we can easily solve. With a solid grasp of this foundational concept, we're well-equipped to tackle the problem at hand.

2. Combining Logarithms: Using Logarithmic Properties

The first step in solving $\log _2(x-5)=3-\log _2(x-7)$ is to consolidate the logarithmic terms. This involves using the properties of logarithms to combine multiple logarithmic expressions into a single one. One of the most important properties for this equation is the sum of logarithms rule: $\log_b(m) + \log_b(n) = \log_b(mn)$. This rule states that the logarithm of the product of two numbers is equal to the sum of their logarithms, provided they have the same base.

In our case, we have a $-\log_2(x-7)$ on the right side of the equation. To use the sum of logarithms property, we need to move this term to the left side. Adding $\log_2(x-7)$ to both sides of the equation gives us: $\log_2(x-5) + \log_2(x-7) = 3$. Now, we have two logarithms with the same base (2) added together on one side of the equation. We can now apply the sum of logarithms property.

Applying the property, we combine the two logarithms into a single one: $\log_2((x-5)(x-7)) = 3$. This simplifies the equation significantly. We've transformed the sum of two logarithms into the logarithm of a product. This is a crucial step because it reduces the complexity of the equation, making it easier to work with. Remember, the goal here is to simplify the equation using the fundamental properties of logarithms. By combining the logarithmic terms, we've paved the way for the next step: converting the logarithmic equation into an exponential equation. This process of simplification is a common strategy in solving various types of equations, and it highlights the power of mathematical properties in making complex problems more manageable.

3. Converting to Exponential Form: Unveiling the Equation

Now that we have a single logarithmic expression, $\log_2((x-5)(x-7)) = 3$, the next step is to convert it into its equivalent exponential form. This is where the fundamental relationship between logarithms and exponents comes into play. Remember, the logarithmic equation $\log_b(a) = c$ is equivalent to the exponential equation $b^c = a$. In our case, the base b is 2, the argument a is (x-5)(x-7), and the exponent c is 3.

Applying this conversion, we rewrite the equation as $2^3 = (x-5)(x-7)$. This transformation is a crucial step because it eliminates the logarithm, allowing us to work with a more familiar algebraic equation. The exponential form directly reveals the relationship between the base, the exponent, and the result, making it easier to isolate and solve for x. Simplifying $2^3$ gives us $8 = (x-5)(x-7)$. We've now successfully converted the logarithmic equation into a quadratic equation, which we can solve using standard algebraic techniques.

This conversion highlights the power of understanding the relationship between logarithms and exponents. By switching between these two forms, we can manipulate equations and solve for unknowns. The exponential form provides a direct and clear relationship between the variables, making the equation more accessible. So, remember this key step: converting from logarithmic to exponential form is a common and effective strategy for solving logarithmic equations. It's like unlocking a secret code that reveals the underlying structure of the equation.

4. Solving the Quadratic Equation: Finding Potential Solutions

With our equation now in the form $8 = (x-5)(x-7)$, we have a quadratic equation to solve. To do this, we first need to expand the right side of the equation. Expanding $(x-5)(x-7)$ gives us $x^2 - 7x - 5x + 35$, which simplifies to $x^2 - 12x + 35$. So, our equation becomes $8 = x^2 - 12x + 35$.

Next, we need to set the equation to zero, which is the standard form for solving quadratic equations. Subtracting 8 from both sides gives us $0 = x^2 - 12x + 27$. Now, we have a standard quadratic equation in the form $ax^2 + bx + c = 0$, where a = 1, b = -12, and c = 27. There are several ways to solve quadratic equations, including factoring, using the quadratic formula, or completing the square. In this case, factoring is a straightforward approach.

We need to find two numbers that multiply to 27 and add up to -12. These numbers are -3 and -9. Therefore, we can factor the quadratic as $(x-3)(x-9) = 0$. Setting each factor to zero gives us two potential solutions: $x-3 = 0$ or $x-9 = 0$. Solving these equations gives us $x = 3$ and $x = 9$. These are our potential solutions, but we're not quite done yet. We need to check these solutions to make sure they are valid in the original logarithmic equation. Remember, the argument of a logarithm must be positive, so we need to make sure our solutions don't violate this condition.

5. Checking for Extraneous Solutions: Ensuring Validity

We've arrived at two potential solutions: $x = 3$ and $x = 9$. However, not all solutions obtained algebraically are necessarily valid solutions to the original logarithmic equation. This is because the domain of logarithmic functions is restricted to positive arguments. We must check for extraneous solutions, which are solutions that satisfy the transformed equation but not the original equation. This is a critical step in solving logarithmic equations.

Let's start by checking $x = 3$. Plugging this value into the original equation, $\log _2(x-5)=3-\log _2(x-7)$, we get $\log_2(3-5) = 3 - \log_2(3-7)$. This simplifies to $\log_2(-2) = 3 - \log_2(-4)$. Notice that we have logarithms of negative numbers. Since the argument of a logarithm must be positive, $x = 3$ is an extraneous solution and is not a valid solution to the equation.

Now, let's check $x = 9$. Plugging this value into the original equation, we get $\log_2(9-5) = 3 - \log_2(9-7)$. This simplifies to $\log_2(4) = 3 - \log_2(2)$. We know that $\log_2(4) = 2$ (since $2^2 = 4$) and $\log_2(2) = 1$ (since $2^1 = 2$). So, the equation becomes $2 = 3 - 1$, which simplifies to $2 = 2$. This is a true statement, so $x = 9$ is a valid solution.

Therefore, after checking for extraneous solutions, we find that only $x = 9$ is a valid solution to the original logarithmic equation. This step highlights the importance of always verifying solutions when dealing with logarithmic equations. Extraneous solutions can arise due to the restricted domain of logarithmic functions, so checking is a crucial part of the solution process.

6. Final Answer: The Solution to the Puzzle

After carefully navigating through the steps of solving the logarithmic equation $\log _2(x-5)=3-\log _2(x-7)$, we've arrived at our final answer. We started by understanding the basics of logarithms, then used logarithmic properties to combine terms. We converted the equation to exponential form, solved the resulting quadratic equation, and, most importantly, checked for extraneous solutions.

Through this process, we found two potential solutions: $x = 3$ and $x = 9$. However, after plugging these values back into the original equation, we discovered that $x = 3$ leads to logarithms of negative numbers, making it an extraneous solution. Only $x = 9$ satisfies the original equation without violating the domain restrictions of logarithms.

Therefore, the solution to the logarithmic equation $\log _2(x-5)=3-\log _2(x-7)$ is $x = 9$. This journey through the equation highlights the importance of a systematic approach to solving mathematical problems. Each step, from understanding the fundamentals to checking for extraneous solutions, plays a crucial role in arriving at the correct answer. So, pat yourselves on the back, guys! You've successfully solved a logarithmic equation!