Solving Logarithmic Equations A Step By Step Guide For Log₅(x+30)=3
In this article, we will delve into the process of solving the logarithmic equation log₅(x+30)=3. Logarithmic equations are a fundamental part of mathematics, particularly in algebra and calculus. Understanding how to solve them is crucial for various applications in science, engineering, and finance. Our main keyword for this discussion is solving logarithmic equations, which we will explore in detail. This article aims to provide a comprehensive, human-friendly guide, ensuring that each step is clear and easy to follow. We will cover the basic principles of logarithms, the transformation of logarithmic equations into exponential forms, and the algebraic manipulations required to isolate the variable x. By the end of this article, you will have a solid understanding of how to tackle similar problems and a deeper appreciation for the elegance of logarithmic functions.
Before we tackle the equation log₅(x+30)=3, let's first make sure we all understand logarithms. A logarithm is essentially the inverse operation to exponentiation. That might sound complicated, but it’s actually quite straightforward. Think of it this way: if we have an exponential equation like b**y = x, the logarithm answers the question, "To what power must we raise b to get x?" This brings us to our main keyword again: logarithmic equations. In mathematical notation, this question is answered by the logarithmic equation logb(x) = y. Here, b is the base of the logarithm, x is the argument, and y is the exponent.
To illustrate this, consider the equation 2³ = 8. Here, 2 is the base, 3 is the exponent, and 8 is the result. The equivalent logarithmic form is log₂(8) = 3. This tells us that we need to raise 2 to the power of 3 to get 8. Understanding this relationship between exponentiation and logarithms is crucial. The logarithmic function undoes the exponential function, and vice versa. This inverse relationship is the key to solving logarithmic equations. One way to ensure that you've grasped the basics is to practice converting between exponential and logarithmic forms. This will give you a foundational understanding that simplifies the more complex steps we'll encounter later. For example, 10² = 100 can be written as log₁₀(100) = 2, and 5² = 25 becomes log₅(25) = 2. This simple conversion exercise reinforces the fundamental concept of logarithms.
The base of a logarithm is a critical component. While any positive number (except 1) can serve as a base, two bases are particularly common and important: base 10 and base e. Base 10 logarithms, often written as log(x) without explicitly mentioning the base, are called common logarithms. Base e logarithms, where e is the mathematical constant approximately equal to 2.71828, are called natural logarithms and are denoted as ln(x). Natural logarithms are incredibly important in calculus and advanced mathematics due to their unique properties related to exponential growth and decay. It’s important to recognize these different notations and understand that they simply indicate different bases.
Another essential aspect of logarithms is their properties. Logarithms have several properties that make them incredibly useful for simplifying complex expressions and solving equations. These include the product rule, the quotient rule, and the power rule. The product rule states that logb(mn) = logb(m) + logb(n), which means the logarithm of a product is the sum of the logarithms. The quotient rule states that logb(m/ n) = logb(m) - logb(n), meaning the logarithm of a quotient is the difference of the logarithms. The power rule states that logb(mp) = p logb(m), which indicates that the logarithm of a number raised to a power is the power times the logarithm of the number. These rules are instrumental in manipulating logarithmic expressions and solving equations. Familiarizing yourself with these properties will make the process of solving logarithmic equations much more manageable.
Now that we have a solid understanding of logarithms, the critical step in solving the equation log₅(x+30)=3 is to convert it into its exponential form. This transformation is the key to unwrapping the logarithm and isolating x. Remember, the logarithmic equation logb(x) = y is equivalent to the exponential equation b**y = x. This is the fundamental relationship we discussed earlier, and it’s the cornerstone of solving logarithmic equations. Understanding this conversion process is essential. Our main keyword continues to be solving logarithmic equations, and this conversion step is a primary technique in that process.
In our specific equation, log₅(x+30)=3, the base b is 5, the argument x is (x+30), and the exponent y is 3. Applying the conversion rule, we rewrite the equation in exponential form as 5³ = x+30. Notice how the logarithm disappears, and we now have a simple algebraic equation to solve. This transformation makes the problem much more approachable. Instead of dealing with a logarithm, we are now working with a straightforward exponentiation and addition. Mastering this conversion technique is like unlocking a door; it allows you to move from the world of logarithms to the more familiar terrain of basic algebra.
To reinforce this conversion, let’s consider a few more examples. Suppose we have log₂(16) = 4. Converting this to exponential form gives us 2⁴ = 16. Another example: log₁₀(1000) = 3 becomes 10³ = 1000. The key is to identify the base, the argument, and the exponent, and then rearrange them into the exponential form. Practice with different examples will help you internalize this process, making it second nature. This skill is not just useful for solving equations; it also helps in understanding the relationship between logarithmic and exponential functions more deeply.
The reason this conversion is so effective is that it allows us to apply the rules of algebra to solve for the variable. Once we have the equation in exponential form, we can use standard algebraic techniques, such as isolating the variable, to find the solution. In the case of 5³ = x+30, we can first evaluate 5³ to get 125, and then we can subtract 30 from both sides to isolate x. This step-by-step approach makes the solution process clear and manageable. By converting to exponential form, we essentially strip away the logarithmic wrapping and reveal the underlying algebraic structure of the equation.
With the logarithmic equation transformed into its exponential form, 5³ = x+30, the next step is to solve for x. This involves basic algebraic manipulation, a process that becomes much simpler once the logarithm is removed. Our primary keyword, solving logarithmic equations, is now channeled into this algebraic solution phase. We’ve already discussed the importance of converting to exponential form, and now we’ll see how that conversion directly leads to a solution.
First, we evaluate 5³. The cube of 5, or 5 multiplied by itself three times, is 5 * 5 * 5, which equals 125. So, our equation becomes 125 = x+30. This simplification is a critical step, as it reduces the equation to a linear form. Now, to isolate x, we need to get it by itself on one side of the equation. To do this, we subtract 30 from both sides of the equation. This maintains the balance of the equation and moves us closer to our goal.
Subtracting 30 from both sides gives us 125 - 30 = x+30 - 30. Simplifying both sides, we get 95 = x. Thus, the solution to the equation is x = 95. This final step is the culmination of all our previous work. We’ve successfully converted the logarithmic equation to exponential form, simplified it, and isolated x to find the solution. This entire process illustrates the power of converting logarithmic equations into a format that allows for standard algebraic methods to be applied.
To ensure accuracy, it’s always a good practice to check the solution by substituting it back into the original equation. In our case, we substitute x = 95 into the original equation, log₅(x+30)=3, to see if it holds true. So, we have log₅(95+30) = log₅(125). Since 5³ = 125, log₅(125) indeed equals 3. This confirms that our solution x = 95 is correct. Verification is a key part of the problem-solving process, especially in mathematics. It not only ensures that you have the correct answer but also reinforces your understanding of the concepts and techniques used.
In summary, we've successfully solved the logarithmic equation log₅(x+30)=3 by converting it to its exponential form, 5³ = x+30, and then using basic algebraic manipulations to isolate x. The solution we found is x = 95. This detailed walkthrough encapsulates our primary keyword: solving logarithmic equations. Understanding the relationship between logarithmic and exponential forms is crucial for solving such equations. The ability to convert between these forms is a fundamental skill in mathematics, enabling the application of algebraic techniques to solve a wider range of problems.
Throughout this article, we’ve emphasized the importance of a step-by-step approach. First, understanding the basic principles of logarithms is essential. Then, the conversion from logarithmic to exponential form is a pivotal step. Finally, applying algebraic techniques to solve for the variable completes the process. This methodical approach not only leads to the correct solution but also enhances comprehension and retention of the concepts. We also highlighted the importance of checking your solution. Substituting the value back into the original equation ensures that the solution is correct and reinforces understanding.
Logarithmic equations are an integral part of mathematics and find applications in various fields, including science, engineering, and finance. Mastering the techniques to solve them is a valuable skill. The process we’ve outlined here – understanding logarithms, converting to exponential form, solving algebraically, and verifying the solution – is a general method applicable to many logarithmic equations. By practicing these steps with different equations, you can build confidence and proficiency in solving logarithmic problems.
We encourage readers to try solving similar equations to further solidify their understanding. Mathematics is a subject best learned through practice, and the more you engage with these concepts, the more comfortable and confident you will become. Logarithmic equations may seem daunting at first, but with a clear understanding of the underlying principles and a systematic approach, they can be solved effectively. The skills you develop in solving logarithmic equations will undoubtedly be beneficial in more advanced mathematical studies and in various real-world applications.
The correct answer is D. x = 95.
