Solving Logarithmic Equations: Finding Exact Solutions

Nov 16, 2025 by Alex Johnson 55 views

Hey math enthusiasts! Today, we're diving into the world of logarithmic equations. Specifically, we'll tackle an equation, find its solution set, and nail down those exact solutions. This is a classic type of problem, perfect for building your algebra skills. Let's get started!

Understanding the Problem: The Logarithmic Equation

The equation we're working with is: $\log x + \log(x - 15) = \log(x - 63)$ . Our goal is to find the values of x that make this equation true. Remember, when we talk about log without specifying a base, we're generally talking about the common logarithm, which has a base of 10. This means we're looking for an x such that the logarithm of x plus the logarithm of (x-15) equals the logarithm of (x-63). It's crucial to understand the properties of logarithms to solve this type of problem effectively.

Let's break down the key concepts before diving into the solution. The logarithm is essentially the inverse function of exponentiation. If we have $\log_b a = c$ , this means $b^c = a$ . Logarithms follow some important rules that will be essential in our problem-solving process. One critical rule is the product rule of logarithms: $\log_b m + \log_b n = \log_b (m \cdot n)$ . This rule lets us combine two separate logarithms into one, greatly simplifying our equation.

In our particular problem, we have addition of two logarithms on one side of the equation and a single logarithm on the other side. By applying the product rule, we can simplify the equation and combine the two logarithms on the left-hand side into a single one. This simplification will lead us closer to finding the solution. Understanding these fundamental rules will help you feel more confident about solving logarithmic equations in the future. Don't worry if it seems a bit tricky at first; with practice, it will all become clear! Remember that the most important thing is to be consistent when you're working with logarithmic functions.

Applying Logarithmic Properties: Simplification is Key

Now, let's use the product rule of logarithms to simplify the given equation: $\log x + \log(x - 15) = \log(x - 63)$ . The product rule tells us that the sum of two logarithms with the same base is equal to the logarithm of the product of their arguments. Applying this to our equation, we get $\log(x \cdot (x - 15)) = \log(x - 63)$ . This step is extremely valuable because it simplifies the equation and allows us to get rid of the logarithms, thus, it significantly simplifies the problem. By applying this rule, we have reduced the complexity of our initial equation. This simplification will bring us closer to isolating x and finding the solution. Without applying this rule, the problem would be significantly more complex and harder to solve.

At this point, we have $\log(x^2 - 15x) = \log(x - 63)$ . Because the logarithms on both sides of the equation have the same base, we can equate the arguments. The argument of a logarithm is the value inside the parentheses. So, we now have $x^2 - 15x = x - 63$ . This simplifies things considerably, transforming our logarithmic equation into a quadratic equation, which we can solve using standard algebraic methods. This is an essential step, it significantly simplifies the equation, which allows us to find x. The key here is realizing that when the logs have the same base, the arguments must be equal for the equation to hold true.

The transformation from a logarithmic equation to a quadratic equation is a common strategy in algebra. It helps us apply different techniques and methods that will help us to find the solution to the equation more quickly and reliably.

Solving the Quadratic Equation: Finding Potential Solutions

We've transformed our equation into $x^2 - 15x = x - 63$ . To solve this, let's rearrange it into the standard quadratic form: $ax^2 + bx + c = 0$ . Subtracting x and adding 63 to both sides, we get $x^2 - 16x + 63 = 0$ . Now we have a simple quadratic equation that we can solve. The most straightforward method is usually to try factoring the quadratic expression, but we can also use the quadratic formula if factoring is tricky.

Let’s try to factor the quadratic expression $x^2 - 16x + 63$ . We're looking for two numbers that multiply to 63 and add up to -16. After a bit of thought, we find that -7 and -9 fit the bill. Therefore, we can factor the quadratic equation into $(x - 7)(x - 9) = 0$ . This means either $(x - 7) = 0$ or $(x - 9) = 0$ . Solving these two simple equations, we get two potential solutions: $x = 7$ and $x = 9$ . We have now found two potential solutions to our initial logarithmic equation. However, it's crucial to check these solutions in the original equation to ensure they are valid. This is an important step when solving logarithmic equations.

Remember, not all solutions to the simplified equation will necessarily be valid solutions to the original logarithmic equation. This is because logarithms are only defined for positive arguments. We have to verify if the values of x that we found are suitable for the original logarithmic equation or not. If a potential solution causes any of the arguments in the original equation to be negative or zero, then that solution is extraneous and must be discarded.

Checking for Extraneous Solutions: The Importance of Validation

Now, we need to check if our potential solutions, $x = 7$ and $x = 9$ , are valid. The arguments of the logarithms in the original equation were x, (x - 15), and (x - 63). We must substitute our potential solutions into these expressions and make sure that each argument is positive. This process is crucial because logarithms are only defined for positive numbers.

Let’s start with $x = 7$ . Substituting this value into the original equation, we get:

$\log 7 + \log(7 - 15) = \log(7 - 63)$

$\log 7 + \log(-8) = \log(-56)$

As we can see, we have logarithms of negative numbers. The logarithm of a negative number is undefined. Therefore, $x = 7$ is not a valid solution and must be discarded. This highlights the importance of checking our answers. Without this verification, we could potentially provide incorrect answers. Always remember to check your work when solving logarithmic equations, especially when simplifying equations and working with different forms of equations.

Next, let’s check $x = 9$ . Substituting this value into the original equation, we have:

$\log 9 + \log(9 - 15) = \log(9 - 63)$

$\log 9 + \log(-6) = \log(-54)$

Here, we also have logarithms of negative numbers, so $x = 9$ is also not a valid solution. Both potential solutions result in the logarithm of a negative number, meaning neither satisfy the conditions of the original logarithmic equation. Remember, always double-check the potential solutions you find, and if you realize there are no solutions that satisfy the original equation, you can discard them.

The Solution Set: What Does It Mean?

Since both $x = 7$ and $x = 9$ are extraneous solutions, meaning they do not satisfy the original equation, the equation has no valid solutions. In mathematics, the solution set is the set of all values that satisfy the equation. In this case, there are no such values. The solution set for this equation is an empty set, often denoted by {} or ∅. This indicates that there are no real numbers that can be plugged into the original equation to make it true. This result is perfectly acceptable, and it highlights the importance of thorough checking.

In this example, the absence of solutions is due to the restrictions placed on the domain of logarithmic functions. The arguments of logarithms must always be positive. If we had skipped the step of verifying our solutions, we would have incorrectly assumed that we had found valid answers, which highlights how important it is to be thorough when working with logarithmic equations. Therefore, the correct answer is A. There is no solution, {}.

Conclusion: Mastering Logarithmic Equations

Congratulations! We've successfully navigated the process of solving a logarithmic equation. We applied the properties of logarithms, solved the resulting quadratic equation, and critically validated our potential solutions to ensure they were valid. This process helps you understand the intricacies of these equations, reinforcing the need to understand both the steps and the properties of logarithms.

Remember these key takeaways:

Understand Logarithmic Properties: The product rule and other properties are essential. Without them, it would be almost impossible to solve these problems.
Simplify, Simplify, Simplify: Use the properties of logarithms to simplify the equation. The easier the equation is, the easier it will be to find the solutions.
Solve with Confidence: Once simplified, apply your algebraic skills to solve the resulting equation.
Always Validate: Check your solutions in the original equation to avoid extraneous solutions.

By following these steps, you'll be well-equipped to solve a wide variety of logarithmic equations. Keep practicing, and you will become more skilled at these types of problems. Each exercise will help solidify your knowledge of logarithms and your ability to solve equations.

External Link: If you want to dive deeper into the world of logarithms, I recommend checking out Khan Academy's Logarithms and Exponentials. They offer some great resources!