Solving Logarithmic Equations Find The Solution Of Log3(x^2 + 6x) = Log3(2x + 12)

In the realm of mathematics, solving equations is a fundamental skill. Among the various types of equations, logarithmic equations hold a significant place. These equations involve logarithms, which are the inverse operations of exponentiation. Mastering the techniques to solve logarithmic equations is crucial for success in algebra, calculus, and other advanced mathematical fields. In this article, we will delve into a specific logarithmic equation and provide a step-by-step guide to finding its solution. We'll explore the properties of logarithms, the process of simplification, and the importance of checking for extraneous solutions. By the end of this guide, you'll be equipped with the knowledge and skills to confidently tackle similar logarithmic equations.

The given equation is:

$\log _3(x^2 + 6x) = \log _3(2x + 12)$

This equation presents a classic example of a logarithmic equation. Our goal is to isolate the variable x and determine the value(s) that satisfy the equation. To achieve this, we'll employ the properties of logarithms and algebraic manipulation. Remember, when dealing with logarithms, it's essential to be mindful of the domain of the logarithmic function, which requires the argument of the logarithm to be positive. We'll keep this in mind as we proceed with the solution.

Understanding Logarithmic Equations

Before we dive into solving the equation, let's establish a clear understanding of logarithmic equations. A logarithmic equation is an equation that contains a logarithm. The logarithm is a mathematical function that determines the exponent to which a base must be raised to produce a given number. The general form of a logarithmic equation is:

$\log_b(x) = y$

where:

• b is the base of the logarithm (a positive number not equal to 1)
• x is the argument of the logarithm (a positive number)
• y is the logarithm itself

The equation $\log_b(x) = y$ is equivalent to the exponential equation $b^y = x$. This relationship between logarithms and exponents is fundamental to solving logarithmic equations. Understanding this connection allows us to transform logarithmic equations into exponential equations, which are often easier to solve.

Logarithmic equations arise in various applications, including compound interest calculations, pH determination in chemistry, and measuring earthquake intensity on the Richter scale. Their versatility makes them a vital tool in many scientific and engineering disciplines.

Step 1: Simplify the Equation

Our first step in solving the equation $\log _3(x^2 + 6x) = \log _3(2x + 12)$ is to simplify it. A key property of logarithms states that if $\log_b(a) = \log_b(c)$, then $a = c$, provided that a and c are positive. This property allows us to eliminate the logarithms from both sides of the equation, simplifying the problem significantly.

Applying this property to our equation, we get:

$x^2 + 6x = 2x + 12$

This simplification transforms the logarithmic equation into a quadratic equation, which we can solve using standard algebraic techniques. The transformation highlights the power of logarithmic properties in simplifying complex equations.

Step 2: Rearrange into a Quadratic Equation

Now that we've eliminated the logarithms, we have a quadratic equation. To solve this equation, we need to rearrange it into the standard quadratic form:

$ax^2 + bx + c = 0$

where a, b, and c are constants.

To rearrange our equation $x^2 + 6x = 2x + 12$, we subtract $2x$ and $12$ from both sides, resulting in:

$x^2 + 6x - 2x - 12 = 0$

Combining like terms, we get:

$x^2 + 4x - 12 = 0$

This is now a standard quadratic equation, which we can solve using factoring, completing the square, or the quadratic formula.

Step 3: Solve the Quadratic Equation

We have the quadratic equation $x^2 + 4x - 12 = 0$. Let's solve it by factoring. We need to find two numbers that multiply to -12 and add up to 4. These numbers are 6 and -2.

Therefore, we can factor the quadratic equation as follows:

$(x + 6)(x - 2) = 0$

To find the solutions, we set each factor equal to zero:

$x + 6 = 0$ or $x - 2 = 0$

Solving for x in each equation, we get:

$x = -6$ or $x = 2$

These are the potential solutions to our equation. However, we must remember the domain restrictions of logarithms and check for extraneous solutions.

Step 4: Check for Extraneous Solutions

Extraneous solutions are solutions that satisfy the transformed equation (in this case, the quadratic equation) but do not satisfy the original logarithmic equation. This occurs because the domain of a logarithmic function is restricted to positive numbers. Therefore, we must check each potential solution in the original equation to ensure that the arguments of the logarithms are positive.

Let's check $x = -6$:

• Argument of the first logarithm: $x^2 + 6x = (-6)^2 + 6(-6) = 36 - 36 = 0$
• Argument of the second logarithm: $2x + 12 = 2(-6) + 12 = -12 + 12 = 0$

Since the arguments are zero, which is not positive, $x = -6$ is an extraneous solution.

Now, let's check $x = 2$:

• Argument of the first logarithm: $x^2 + 6x = (2)^2 + 6(2) = 4 + 12 = 16$
• Argument of the second logarithm: $2x + 12 = 2(2) + 12 = 4 + 12 = 16$

Since both arguments are positive, $x = 2$ is a valid solution.

Step 5: State the Solution

After checking for extraneous solutions, we can confidently state the solution to the equation $\log _3(x^2 + 6x) = \log _3(2x + 12)$. The only solution is:

$x = 2$

Conclusion

In this article, we've meticulously walked through the process of solving the logarithmic equation $\log _3(x^2 + 6x) = \log _3(2x + 12)$. We began by understanding the properties of logarithms, then simplified the equation using these properties. We transformed the logarithmic equation into a quadratic equation, which we solved by factoring. Finally, and crucially, we checked for extraneous solutions, ensuring the validity of our answer.

This process highlights the importance of a systematic approach to solving mathematical problems. By understanding the underlying principles, applying appropriate techniques, and carefully checking our work, we can confidently tackle even complex equations. Logarithmic equations are a fundamental part of mathematics, and mastering their solution is an essential skill for any aspiring mathematician or scientist.

Remember, the key takeaways from this exercise are:

• Understanding the properties of logarithms is crucial for simplifying logarithmic equations.
• Transforming logarithmic equations into algebraic equations is a common and effective strategy.
• Checking for extraneous solutions is essential to ensure the validity of the solutions.

With these principles in mind, you'll be well-equipped to solve a wide range of logarithmic equations and excel in your mathematical endeavors. Keep practicing, and you'll find that solving these equations becomes second nature. The beauty of mathematics lies in its logical structure and the satisfaction of finding solutions to challenging problems.

To further enhance your problem-solving skills with logarithmic equations, consider these additional tips:

• Practice regularly: The more you practice, the more comfortable you'll become with the various techniques and nuances of solving logarithmic equations. Work through a variety of examples to build your confidence.
• Master the properties of logarithms: A solid understanding of logarithmic properties is essential for simplifying and solving these equations. Review and memorize these properties, and practice applying them in different contexts.
• Be mindful of the domain: Always remember that the argument of a logarithm must be positive. This is crucial for identifying and eliminating extraneous solutions.
• Use different solution methods: Familiarize yourself with various methods for solving quadratic equations (factoring, completing the square, quadratic formula). This will give you flexibility in choosing the most efficient method for a particular problem.
• Check your work: Always double-check your solutions in the original equation to ensure accuracy. This is especially important when dealing with extraneous solutions.

By incorporating these tips into your problem-solving approach, you'll significantly improve your ability to solve logarithmic equations effectively and efficiently. Mathematics is a journey of continuous learning and improvement, so embrace the challenge and enjoy the process of discovery.