Solving Logarithmic Equations A Step By Step Guide
In this comprehensive guide, we will delve into the process of solving the logarithmic equation $\log (x^2-15)=\log (2 x)$. Logarithmic equations, while seemingly complex, can be systematically solved by following a series of well-defined steps. This article will not only provide you with the solution but also equip you with a deep understanding of the underlying principles involved. We'll break down each step with detailed explanations, ensuring that you grasp the nuances of logarithmic equations and their solutions.
1. Equating the Arguments: Setting the Stage for Solving Logarithmic Equations
The cornerstone of solving this equation, $\log (x^2-15)=\log (2 x)$, lies in understanding a fundamental property of logarithms: if $\log_b a = \log_b c$, then $a = c$. This property allows us to eliminate the logarithms and focus on the algebraic expressions within. The initial move involves recognizing that both sides of the equation have the same logarithmic base (which is 10, by default, when no base is explicitly written). This crucial observation allows us to equate the arguments of the logarithms. Therefore, the first step in solving our equation is to set the expressions inside the logarithms equal to each other, which transforms the logarithmic equation into a standard algebraic equation. This simplification is key to unlocking the solution, as it allows us to apply familiar algebraic techniques. The equation we obtain after equating the arguments is $x^2 - 15 = 2x$, a quadratic equation that we can solve using various methods. This step effectively bridges the gap between logarithmic expressions and algebraic manipulations, laying the foundation for finding the values of $x$ that satisfy the original equation. In essence, this initial step is a strategic maneuver that simplifies the problem, making it more accessible and solvable through traditional algebraic means. By understanding the underlying principle of equating arguments, we can confidently navigate the complexities of logarithmic equations and pave the way for a clear and concise solution.
2. Transforming to a Quadratic Equation: The Heart of the Solution
After equating the arguments in the logarithmic equation, $\log (x^2-15)=\log (2 x)$, we arrive at the equation $x^2 - 15 = 2x$. The next crucial step is to transform this equation into the standard form of a quadratic equation. This transformation is essential because it allows us to apply well-established methods for solving quadratic equations, such as factoring, completing the square, or using the quadratic formula. To achieve this standard form, we need to rearrange the terms so that all terms are on one side of the equation, and the other side is zero. This involves subtracting $2x$ from both sides of the equation. By doing so, we obtain the equation $x^2 - 2x - 15 = 0$. This is now a quadratic equation in the form $ax^2 + bx + c = 0$, where $a = 1$, $b = -2$, and $c = -15$. The transformation into this standard form is a pivotal moment in the solution process. It not only simplifies the equation but also opens the door to a range of techniques designed specifically for solving quadratic equations. The ability to recognize and execute this transformation is a fundamental skill in algebra, and it is particularly crucial when dealing with logarithmic equations that often lead to quadratic expressions. This step is not merely a mechanical manipulation; it's a strategic move that aligns the equation with a familiar and solvable form, paving the way for the subsequent steps in finding the solution.
3. Factoring the Quadratic: Unveiling Potential Solutions
Now that we have the quadratic equation in its standard form, $x^2 - 2x - 15 = 0$, the next step is to factor the quadratic expression. Factoring is a powerful technique for solving quadratic equations, as it allows us to rewrite the equation as a product of two binomials. If we can find two binomials that multiply to give the quadratic expression, then we can set each binomial equal to zero and solve for $x$. This is based on the principle that if the product of two factors is zero, then at least one of the factors must be zero. To factor the quadratic $x^2 - 2x - 15$, we need to find two numbers that multiply to $-15$ (the constant term) and add to $-2$ (the coefficient of the $x$ term). These numbers are $-5$ and $3$, since $(-5) imes 3 = -15$ and $(-5) + 3 = -2$. Therefore, we can factor the quadratic as $(x - 5)(x + 3) = 0$. This factorization is a crucial step in the solution process. It transforms the quadratic equation into a form that is easily solvable. By expressing the quadratic as a product of two linear factors, we can apply the zero-product property to find the potential solutions for $x$. Factoring not only simplifies the equation but also provides a clear pathway to identifying the values of $x$ that make the equation true. This step is a testament to the power of algebraic manipulation in unraveling the solutions to complex equations. The ability to factor quadratics efficiently is a valuable skill in mathematics, and it is particularly useful in the context of solving logarithmic and other types of equations.
4. Finding Potential Solutions: Zeroing in on the Answers
With the quadratic equation factored as $(x - 5)(x + 3) = 0$, we can now apply the zero-product property. This property states that if the product of two factors is zero, then at least one of the factors must be zero. In other words, if $AB = 0$, then either $A = 0$ or $B = 0$ (or both). Applying this property to our factored equation, we set each factor equal to zero: $x - 5 = 0$ or $x + 3 = 0$. Solving these two linear equations gives us the potential solutions for $x$. For the first equation, $x - 5 = 0$, we add $5$ to both sides to get $x = 5$. For the second equation, $x + 3 = 0$, we subtract $3$ from both sides to get $x = -3$. Therefore, the potential solutions to the quadratic equation are $x = 5$ and $x = -3$. These values are potential solutions because they satisfy the factored quadratic equation. However, it is crucial to remember that we started with a logarithmic equation, and not all solutions to the quadratic equation will necessarily be solutions to the original logarithmic equation. This is because the domain of a logarithmic function is restricted to positive arguments. Therefore, we must check these potential solutions in the original logarithmic equation to ensure that they do not lead to logarithms of negative numbers or zero. Finding these potential solutions is a significant step in the solution process. It narrows down the possibilities and provides us with the candidate values of $x$ that might satisfy the original equation. However, the next step is equally important: verifying these solutions to ensure their validity in the context of the original logarithmic equation.
5. Verifying Solutions: Ensuring Validity in Logarithmic Equations
After finding the potential solutions $x = 5$ and $x = -3$, the final and arguably most critical step is to verify these solutions in the original logarithmic equation, $\log (x^2-15)=\log (2 x)$. This step is essential because logarithmic functions have a restricted domain: the argument of a logarithm must be strictly positive. Therefore, any potential solution that results in a non-positive argument for any logarithm in the equation is an extraneous solution and must be discarded. Let's start by checking $x = 5$. Substituting $x = 5$ into the original equation, we get: $\log (5^2 - 15) = \log (25 - 15) = \log (10)$. And $\& \log (2 imes 5) = \log (10)$. Since both logarithms are defined and equal, $x = 5$ is a valid solution. Now, let's check $x = -3$. Substituting $x = -3$ into the original equation, we get: $\log ((-3)^2 - 15) = \log (9 - 15) = \log (-6)$. Since the argument of the logarithm is negative, this expression is undefined. Therefore, $x = -3$ is not a valid solution. This verification step highlights the importance of considering the domain restrictions of logarithmic functions. Even if a value satisfies the transformed algebraic equation, it may not be a solution to the original logarithmic equation if it leads to the logarithm of a non-positive number. In conclusion, after careful verification, we find that the only valid solution to the equation $\log (x^2-15)=\log (2 x)$ is $x = 5$. This comprehensive process of solving logarithmic equations, from equating arguments to verifying solutions, provides a robust framework for tackling similar problems in mathematics.
Therefore, the correct order of steps to solve the equation $\log (x^2-15)=\log (2 x)$ is:
-
-
-
x-5=0$ or $x+3=0
- Potential solutions are -3 and 5
- Verifying Solutions
