Equivalent Equation For Log₃(x+5) = 2: A Step-by-Step Guide

Hey there, math enthusiasts! Today, we're diving into the fascinating world of logarithmic equations. Specifically, we're going to break down the equation log₃(x+5) = 2 and figure out which equivalent equation correctly represents it. This is a common type of problem you'll encounter in algebra and precalculus, so let's get a solid understanding of how to tackle it. Guys, don't worry if logarithms seem a bit intimidating at first; we'll take it step by step and make sure you're confident by the end. We'll explore the fundamental relationship between logarithms and exponents, and how that relationship allows us to rewrite logarithmic equations in exponential form. This is a crucial skill for solving these types of problems, so let's jump right in!

Understanding Logarithms and Exponents

Before we dive into the specific equation, let's quickly review the relationship between logarithms and exponents. This is the key to unlocking logarithmic equations. Think of logarithms as the inverse operation of exponentiation. Just like subtraction undoes addition, and division undoes multiplication, logarithms "undo" exponents. In simpler terms, a logarithm answers the question: "To what power must we raise the base to get a certain number?" For instance, consider the expression logₐ(b) = c. This logarithmic equation is equivalent to the exponential equation aᶜ = b. Here, 'a' is the base, 'c' is the exponent (or power), and 'b' is the result. The logarithm, therefore, tells us that if we raise the base 'a' to the power of 'c', we get 'b'. To really grasp this, let's look at some concrete examples. For instance, log₂(8) = 3 because 2 raised to the power of 3 equals 8 (2³ = 8). Similarly, log₁₀(100) = 2 because 10 squared is 100 (10² = 100). Understanding this fundamental connection is crucial because it allows us to convert between logarithmic and exponential forms, which is often the first step in solving logarithmic equations. So, with this basic principle in mind, let’s circle back to our original equation and see how we can apply this knowledge.

The Importance of the Base

Let's take a moment to stress the significance of the base in both logarithmic and exponential forms. The base is the foundation upon which the entire relationship rests. It's the number that's being raised to a power in the exponential form, and it's the subscript in the logarithmic form. If you change the base, you completely change the relationship. Imagine trying to build a house on a weak foundation – it simply won't stand. Similarly, if you misidentify the base, you'll end up with an incorrect equation. Think about our example, log₂(8) = 3. The base here is 2. If we were to mistakenly use a different base, say 10, we'd have log₁₀(8), which is definitely not equal to 3. The base is an integral part of the equation, and we must treat it with the utmost respect. When converting between logarithmic and exponential forms, make sure you correctly identify and maintain the base. This will save you a lot of headaches down the road. The base is constant. It remains the same whether it is converted to exponential form or remains in logarithmic form.

Applying the Logarithmic-Exponential Relationship to Our Equation

Now that we've refreshed our understanding of the relationship between logarithms and exponents, let's apply it to our equation: log₃(x+5) = 2. Remember the general form: logₐ(b) = c is equivalent to aᶜ = b. In our case, we can identify: a = 3 (the base), b = (x+5) (the argument of the logarithm), and c = 2 (the result of the logarithm). So, to rewrite this logarithmic equation in its equivalent exponential form, we follow the pattern: a raised to the power of c equals b. Substituting our values, we get: 3² = x + 5. This is the crucial step in solving this problem. We've successfully transformed the logarithmic equation into a much simpler exponential equation. Guys, see how the logarithm just melts away when we apply the exponential form? This is the magic of understanding the inverse relationship. Now, let's compare this result to the given options and see which one matches. It's like we're detectives, and this exponential equation is our key piece of evidence!

Common Mistakes to Avoid

Before we move on, let's take a moment to address some common mistakes people often make when dealing with logarithmic equations. Being aware of these pitfalls can save you from making errors. One frequent mistake is confusing the base and the exponent. Remember, the base is the small subscript number in the logarithm, and the exponent is the number the base is raised to. Another error is incorrectly applying the logarithmic-exponential relationship. It's essential to follow the correct pattern: logₐ(b) = c becomes aᶜ = b. Don't mix up the positions of a, b, and c! Also, watch out for algebraic errors when solving the resulting exponential equation. A simple mistake in arithmetic can throw off your entire solution. For example, students sometimes try to apply the exponent to both sides of the logarithmic expression, like raising [log₃(x+5)] to a power, which is not the correct approach for converting the equation. By keeping these potential pitfalls in mind, you can approach logarithmic equations with greater confidence and accuracy.

Identifying the Equivalent Equation

Okay, we've successfully transformed our logarithmic equation log₃(x+5) = 2 into its equivalent exponential form: 3² = x + 5. Now, let's play detective and compare this equation to the options provided. We are looking for the option that exactly matches 3² = x + 5. Let's consider the given options:

• 3² = [log₃(x+5)]³: This is incorrect. We correctly converted the log equation to exponential form. Cubing the log expression is not the correct way to go.
• 2³ = [log₃(x+5)]²: This is incorrect as well. Similarly, this option involves squaring the log expression and changing the base, which is not in line with the correct conversion process.
• 3² = x + 5: Bingo! This is exactly the equation we derived by applying the logarithmic-exponential relationship.
• 2³ = x + 5: This is incorrect. This option seems to confuse the base and the result of the logarithm.

Therefore, the equivalent equation is 3² = x + 5. Guys, see how carefully stepping through the process and understanding the underlying principles makes the answer clear? It's not about guessing; it's about knowing!

Verifying the Solution

It's always a good practice to verify your solution, especially in mathematics. This helps to ensure that you have correctly solved the problem and haven't made any mistakes along the way. In our case, we've identified 3² = x + 5 as the equivalent equation. We can take it a step further and solve for x to see if it makes sense in the original equation. First, simplify 3² = x + 5 to 9 = x + 5. Then, subtract 5 from both sides to isolate x: 9 - 5 = x, which gives us x = 4. Now, let's substitute x = 4 back into the original logarithmic equation: log₃(x+5) = 2. We get log₃(4+5) = log₃(9). And since 3² = 9, log₃(9) = 2. So, our solution checks out! Verifying the solution not only gives you confidence in your answer but also reinforces your understanding of the concepts involved. It's a win-win!

Conclusion: Mastering Logarithmic Equations

Great job, everyone! We've successfully navigated the world of logarithmic equations and determined that 3² = x + 5 is indeed the equation equivalent to log₃(x+5) = 2. The key takeaway here is understanding the fundamental relationship between logarithms and exponents. By remembering that logarithms are the inverse operation of exponentiation, we can confidently convert between logarithmic and exponential forms. This is a powerful tool for solving logarithmic equations. We also emphasized the importance of correctly identifying the base and avoiding common mistakes. And don't forget the value of verifying your solution – it's the ultimate safety net! Guys, with practice and a solid understanding of these principles, you'll be able to tackle any logarithmic equation that comes your way. Keep practicing, keep exploring, and keep those math skills sharp!