Solving Logarithmic Equations A Step-by-Step Guide For Log₃(x² + 6) = 3
Hey there, math enthusiasts! Today, we're diving deep into the fascinating world of logarithms to solve a tricky equation. Logarithmic equations might seem daunting at first, but with a clear understanding of the underlying principles, we can conquer them with ease. So, let's jump right in and crack this mathematical puzzle together!
The Challenge: Deciphering log₃(x² + 6) = 3
Our mission, should we choose to accept it, is to find the value(s) of 'x' that satisfy the equation log₃(x² + 6) = 3. This equation throws a logarithmic curveball our way, but fear not! We're equipped with the right tools to tackle it head-on. The key to solving logarithmic equations lies in understanding the relationship between logarithms and exponents. Remember, logarithms are simply the inverse of exponential functions. In simpler terms, a logarithm answers the question: "What exponent do I need to raise the base to in order to get this number?"
To truly grasp the essence of this equation, let's break it down piece by piece. The logarithm symbol, log₃, tells us that we're dealing with a base-3 logarithm. This means we're asking ourselves, "What power of 3 will give us the value inside the parentheses?" The expression inside the parentheses, (x² + 6), is the argument of the logarithm. It represents the value we're trying to obtain by raising 3 to a certain power. On the right side of the equation, we have the number 3. This is the result of the logarithm, the exponent we're looking for. So, the equation is essentially saying, "3 raised to the power of 3 equals x² + 6." Now that we've dissected the equation, we can see the underlying structure more clearly. It's like deciphering a secret code – once you understand the symbols, the message becomes clear.
The first step in solving this equation is to convert it from logarithmic form to exponential form. This transformation will help us get rid of the logarithm and work with a more familiar algebraic expression. The fundamental relationship between logarithms and exponents is expressed as follows: logₐ(b) = c is equivalent to aᶜ = b. Applying this principle to our equation, where 'a' is 3, 'b' is (x² + 6), and 'c' is 3, we can rewrite the equation as: 3³ = x² + 6. See how we've transformed the logarithmic equation into a simple exponential one? This is a crucial step in solving these types of problems. By understanding this conversion, we've unlocked the door to solving for 'x'. The equation now looks much more manageable, doesn't it? We've effectively translated the problem into a language we can easily work with. From here, the path to the solution becomes clearer and more straightforward.
Exponential Transformation: Unveiling the Algebraic Equation
As we discovered in the previous section, the key to solving our logarithmic equation log₃(x² + 6) = 3 lies in transforming it into its exponential counterpart. By applying the fundamental relationship between logarithms and exponents, we successfully rewrote the equation as 3³ = x² + 6. This transformation is a game-changer because it allows us to work with a standard algebraic equation, which is much easier to manipulate and solve.
Now, let's simplify the exponential term. We know that 3³ (3 raised to the power of 3) is simply 3 * 3 * 3, which equals 27. So, our equation now becomes 27 = x² + 6. This is a quadratic equation in disguise! We've successfully shed the logarithmic cloak and revealed the algebraic nature of the problem. Quadratic equations are a staple in algebra, and we have a variety of tools at our disposal to solve them. Whether it's factoring, completing the square, or using the quadratic formula, we're well-equipped to find the solutions for 'x'.
To prepare for solving the quadratic equation, we need to rearrange it into the standard form, which is ax² + bx + c = 0. In our case, we need to move the constant term (27) to the right side of the equation. We can do this by subtracting 27 from both sides. This gives us: 0 = x² + 6 - 27. Now, let's simplify the right side by combining the constant terms: 0 = x² - 21. Voila! We have successfully transformed the equation into the standard quadratic form. This form makes it easier to identify the coefficients and apply the appropriate solution methods. We've now set the stage for the final act: solving for 'x'. The equation is neatly arranged, and the path to the solution is clear. We're ready to unleash our algebraic prowess and find the values of 'x' that satisfy the original logarithmic equation.
Solving the Quadratic: Finding the Values of x
With our equation now in the standard quadratic form, x² - 21 = 0, we're ready to embark on the final stage of our mathematical journey: solving for 'x'. There are several methods we could employ to tackle this quadratic, but in this case, the simplest and most efficient approach is to use the square root property. This method is particularly well-suited for quadratic equations where the linear term (the term with 'x') is absent, as in our equation.
The square root property states that if x² = k, then x = ±√k, where 'k' is a constant. In essence, we isolate the x² term and then take the square root of both sides, remembering to consider both the positive and negative roots. To apply this property to our equation, we first need to isolate the x² term. We can do this by adding 21 to both sides of the equation: x² = 21. Now, we're perfectly set up to use the square root property.
Taking the square root of both sides, we get: x = ±√21. This gives us two potential solutions: x = √21 and x = -√21. These are the values of 'x' that, when squared and added to 6, will result in a value whose base-3 logarithm is equal to 3. We've successfully navigated the logarithmic and algebraic landscape to arrive at our solutions. However, before we declare victory, it's crucial to verify that these solutions are valid within the context of the original logarithmic equation.
It's important to remember that logarithms are only defined for positive arguments. This means that the expression inside the logarithm, (x² + 6) in our case, must be greater than zero. Let's check if our solutions, x = √21 and x = -√21, satisfy this condition. When x = √21, x² = 21, and x² + 6 = 27, which is indeed positive. Similarly, when x = -√21, x² = 21 (since squaring a negative number results in a positive number), and x² + 6 = 27, which is also positive. Therefore, both solutions are valid and do not violate the domain of the logarithm.
The Grand Finale: Validating the Solutions
We've journeyed through the realm of logarithms, navigated the terrain of quadratic equations, and arrived at two potential solutions: x = √21 and x = -√21. However, our quest isn't complete until we've rigorously validated these solutions against the original equation, log₃(x² + 6) = 3. Validation is a crucial step in solving any mathematical problem, especially those involving logarithms and radicals, where extraneous solutions can sometimes creep in. An extraneous solution is a value that satisfies a transformed equation but not the original one.
To validate our solutions, we'll substitute each value of 'x' back into the original equation and check if the equation holds true. Let's start with x = √21. Substituting this value into the equation, we get: log₃((√21)² + 6) = log₃(21 + 6) = log₃(27). Now, we need to determine if log₃(27) is indeed equal to 3. Recall that logarithms ask the question, "What power do I need to raise the base to in order to get this number?" In this case, we're asking, "What power of 3 gives us 27?" The answer is 3, since 3³ = 27. Therefore, log₃(27) = 3, and the equation holds true for x = √21.
Now, let's validate the second solution, x = -√21. Substituting this value into the equation, we get: log₃((-√21)² + 6) = log₃(21 + 6) = log₃(27). Notice that squaring a negative number also results in a positive number, so we end up with the same expression as before. Again, we know that log₃(27) = 3, so the equation holds true for x = -√21 as well.
Both of our solutions have passed the validation test! This confirms that they are indeed the true solutions to the original logarithmic equation. We've successfully navigated the entire problem-solving process, from transforming the equation to validating the results. This comprehensive approach ensures that we not only find potential solutions but also verify their correctness, leaving no room for doubt.
Conclusion: Triumphantly Solving the Logarithmic Puzzle
We've reached the summit of our mathematical expedition, and what a journey it has been! We set out to solve the logarithmic equation log₃(x² + 6) = 3, and through a combination of logarithmic transformations, algebraic manipulations, and rigorous validation, we've successfully unearthed the solutions. Our efforts have revealed that the values of 'x' that satisfy the equation are x = √21 and x = -√21.
This problem serves as a testament to the power of understanding fundamental mathematical principles. By grasping the relationship between logarithms and exponents, we were able to transform the equation into a more manageable form. We then employed our algebraic skills to solve the resulting quadratic equation, and finally, we validated our solutions to ensure their accuracy.
The process of solving this equation has not only provided us with the answer but has also reinforced our understanding of key mathematical concepts. We've seen how logarithms and exponents are intertwined, how quadratic equations can arise in unexpected places, and how crucial validation is in the problem-solving process. These are valuable lessons that will serve us well in future mathematical endeavors.
So, the next time you encounter a logarithmic equation, remember the steps we've taken today. Transform, simplify, solve, and validate. With these tools in your arsenal, you'll be well-equipped to conquer any mathematical challenge that comes your way. Keep exploring, keep learning, and keep unlocking the mysteries of mathematics!
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