Solving Log(x^2-15)=log(2x) Equation A Step-by-Step Guide

Logarithmic equations can seem daunting at first, but by following a systematic approach, they become much more manageable. This guide will walk you through the steps to solve a logarithmic equation, using the example equation $\log (x^2 - 15) = \log (2x)$. We'll break down each step and explain the reasoning behind it, ensuring you have a solid understanding of the process. This comprehensive guide aims to provide a clear and concise method for tackling logarithmic equations, enhancing your problem-solving skills and confidence in mathematics. Understanding the underlying principles of logarithms is crucial for mastering these equations. Logarithms are essentially the inverse operation of exponentiation, and this relationship is key to solving logarithmic problems. By converting logarithmic equations into their exponential form, we can often simplify the equation and make it easier to solve. In addition to understanding the relationship between logarithms and exponents, it's also important to be familiar with the properties of logarithms, such as the product rule, quotient rule, and power rule. These properties allow us to manipulate logarithmic expressions and combine or separate terms, which can be very helpful in solving equations. This article will not only guide you through the specific steps for solving the given equation but also equip you with the broader knowledge and skills needed to tackle a variety of logarithmic problems. Remember, practice is key to mastering any mathematical concept, so work through several examples and don't hesitate to seek help when needed.

Step 1: Equate the Arguments

The initial step in solving the equation $\log (x^2 - 15) = \log (2x)$ involves recognizing that if the logarithms of two expressions are equal, then the expressions themselves must be equal. This principle stems from the fact that the logarithmic function is a one-to-one function, meaning that each input has a unique output, and vice versa. Therefore, if $\log_b(A) = \log_b(B)$, then it must be true that $A = B$. Applying this principle to our equation, we can eliminate the logarithms and set the arguments equal to each other. This simplifies the equation and allows us to work with a more familiar algebraic form. Specifically, we set $x^2 - 15$ equal to $2x$, which results in the equation $x^2 - 15 = 2x$. This step is crucial because it transforms a logarithmic equation into a standard quadratic equation, which we can then solve using well-established methods. The ability to make this transformation is a fundamental skill in solving logarithmic equations. It's important to remember that this step is only valid if the bases of the logarithms are the same. In this case, since the base is not explicitly written, it is assumed to be 10 (the common logarithm), and the bases are indeed the same. However, if the bases were different, we would need to use other techniques to solve the equation. By understanding this principle, you can confidently apply it to a wide range of logarithmic equations, making the solving process more efficient and accurate. This step sets the stage for the subsequent steps in the solution, ultimately leading to the identification of potential solutions for x.

Step 2: Form a Quadratic Equation

After equating the arguments, we have the equation $x^2 - 15 = 2x$. To solve for x, the next step is to rearrange this equation into the standard form of a quadratic equation, which is $ax^2 + bx + c = 0$. This form is essential because it allows us to use various methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. To achieve this standard form, we need to move all terms to one side of the equation, leaving zero on the other side. In this case, we subtract $2x$ from both sides of the equation $x^2 - 15 = 2x$. This operation maintains the equality of the equation while bringing all terms to the left-hand side. The resulting equation is $x^2 - 2x - 15 = 0$. This equation is now in the standard quadratic form, where $a = 1$, $b = -2$, and $c = -15$. Having the equation in this form is a significant step forward because it opens up a range of techniques for finding the solutions. Factoring, if possible, is often the quickest method. If factoring is not straightforward, the quadratic formula or completing the square can be used. The importance of this step cannot be overstated; it transforms a logarithmic problem into a more familiar algebraic problem, making it accessible to a wider range of solution methods. By mastering this step, you'll be well-equipped to tackle a variety of logarithmic and algebraic problems.

Step 3: Factor the Quadratic Equation

With the quadratic equation in the standard form $x^2 - 2x - 15 = 0$, we now aim to solve for x. One of the most efficient methods for solving quadratic equations is factoring, if it's possible. Factoring involves expressing the quadratic expression as a product of two binomials. To factor $x^2 - 2x - 15$, we look for two numbers that multiply to -15 (the constant term) and add up to -2 (the coefficient of the x term). These two numbers are -5 and 3, since (-5) * 3 = -15 and (-5) + 3 = -2. Therefore, we can factor the quadratic equation as $(x - 5)(x + 3) = 0$. This factorization is a crucial step because it transforms the quadratic equation into a product of two factors that equal zero. According to the zero-product property, if the product of two factors is zero, then at least one of the factors must be zero. This property allows us to set each factor equal to zero and solve for x separately. Factoring is a powerful technique for solving quadratic equations, but it's not always possible. If a quadratic equation cannot be factored easily, alternative methods such as the quadratic formula or completing the square can be used. However, when factoring is feasible, it's often the quickest and most straightforward approach. By mastering factoring techniques, you'll be able to solve a wide range of quadratic equations efficiently. This step is a key component in the overall process of solving the logarithmic equation, as it provides us with potential solutions for x.

Step 4: Solve for x

Having factored the quadratic equation as $(x - 5)(x + 3) = 0$, we now apply 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. This leads us to set each factor equal to zero and solve for x. First, we set $x - 5 = 0$, which gives us $x = 5$. Second, we set $x + 3 = 0$, which gives us $x = -3$. Therefore, the potential solutions for the equation are $x = 5$ and $x = -3$. These values are potential solutions because they satisfy the factored quadratic equation. However, it's crucial to remember that we started with a logarithmic equation, and logarithmic functions have domain restrictions. Specifically, the argument of a logarithm must be positive. This means that we need to check whether these potential solutions are valid in the original logarithmic equation. This step is a critical part of the solving process because it ensures that we only accept solutions that are mathematically sound. Failing to check the solutions can lead to incorrect answers, as some potential solutions may not be in the domain of the logarithmic function. By carefully solving for x and then verifying the solutions, we can confidently determine the correct answer to the logarithmic equation.

Step 5: Check for Extraneous Solutions

The final, and arguably one of the most important, steps in solving logarithmic equations is to check for extraneous solutions. Extraneous solutions are potential solutions that satisfy the transformed equation (in this case, the quadratic equation) but do not satisfy the original logarithmic equation. These solutions arise because the domain of logarithmic functions is restricted to positive arguments. To check for extraneous solutions, we must substitute each potential solution back into the original logarithmic equation, $\log (x^2 - 15) = \log (2x)$, and verify that the arguments of the logarithms are positive. Let's first check $x = 5$. Substituting $x = 5$ into the equation, we get $\log (5^2 - 15) = \log (25 - 15) = \log (10)$ and $\log (2 * 5) = \log (10)$. Since both arguments are positive (10 > 0), $x = 5$ is a valid solution. Next, let's check $x = -3$. Substituting $x = -3$ into the equation, we get $\log ((-3)^2 - 15) = \log (9 - 15) = \log (-6)$. Since the argument -6 is negative, $x = -3$ is an extraneous solution and must be discarded. The other side of the equation gives us $\log (2 * -3) = \log (-6)$, which also has a negative argument. Therefore, the only valid solution to the equation $\log (x^2 - 15) = \log (2x)$ is $x = 5$. Checking for extraneous solutions is a critical step that ensures the accuracy of your answer. It's a common mistake to skip this step, but it can lead to incorrect solutions. By carefully checking each potential solution, you can have confidence in your answer and avoid errors. This final check reinforces the importance of understanding the domain restrictions of logarithmic functions and applying them consistently throughout the solving process.

Ordered Steps

Based on the step-by-step explanation above, the correct order of the steps to solve the equation $\log(x^2 - 15) = \log(2x)$ is:

1. $x^2 - 15 = 2x$
2. $x^2 - 2x - 15 = 0$
3. $x-5=0$ or $x+3=0$
4. Potential solutions are -3 and 5

Conclusion

Solving logarithmic equations requires a systematic approach, and this guide has outlined the key steps involved. By understanding the underlying principles of logarithms, equating arguments, forming and factoring quadratic equations, solving for potential solutions, and, most importantly, checking for extraneous solutions, you can confidently tackle a wide range of logarithmic problems. Remember to always be mindful of the domain restrictions of logarithmic functions and to practice consistently to improve your skills. With a clear understanding of these steps, you'll be well-equipped to solve even the most challenging logarithmic equations.