Solving Logarithmic Equations: Find X In Log₂(x+3)=10

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Hey guys! Let's dive into solving a logarithmic equation today. We're going to tackle the equation log₂(x+3) = 10. Logarithmic equations might seem intimidating at first, but don't worry, we'll break it down step by step. By the end of this guide, you'll not only know how to solve this specific problem but also understand the general principles behind solving logarithmic equations. We'll make sure to cover the key concepts and provide clear explanations so you can confidently handle similar problems in the future. Let's get started and unlock the secrets of logarithms together!

Understanding Logarithms

Before we jump into solving the equation, let's make sure we're all on the same page about what a logarithm actually is. At its heart, a logarithm is the inverse operation to exponentiation. Think of it this way: if 2 raised to the power of 3 equals 8 (2³ = 8), then the logarithm base 2 of 8 is 3 (log₂8 = 3). The logarithm asks the question: "What exponent do I need to raise the base to in order to get this number?" This is a crucial concept to grasp, as it forms the foundation for everything we'll be doing next. We can generalize this idea with the following relationship: if bˣ = y, then log_b(y) = x. Here, 'b' is the base of the logarithm, 'x' is the exponent, and 'y' is the result. Understanding this relationship is key to converting between logarithmic and exponential forms, which is a technique we'll use extensively in solving our equation. The base of the logarithm tells us which number is being raised to a power, and the logarithm itself tells us what that power is. It's like a secret code that unlocks the relationship between exponents and their results. Once you're comfortable with this fundamental concept, solving logarithmic equations becomes much more manageable. So, keep this definition in mind as we proceed, and you'll see how it all comes together.

Converting Logarithmic to Exponential Form

Now that we've refreshed our understanding of logarithms, let's apply this knowledge to our specific equation: log₂(x+3) = 10. The first key step in solving this equation is to convert it from logarithmic form into its equivalent exponential form. This conversion allows us to get rid of the logarithm and work with a more familiar algebraic structure. Remember our earlier definition: if log_b(y) = x, then bˣ = y. We can directly apply this relationship to our equation. Here, the base 'b' is 2, the exponent 'x' is 10, and 'y' is (x+3). So, applying the conversion, we rewrite log₂(x+3) = 10 as 2¹⁰ = x+3. See how we've transformed the equation? The logarithm is gone, and we now have a simple exponential equation to work with. This is a crucial step in solving logarithmic equations because it allows us to use the properties of exponents and algebra to isolate the variable we're trying to find. Converting to exponential form is like unlocking a door to a new way of looking at the problem, making it much easier to solve. Take a moment to appreciate this transformation – it's the magic trick that makes logarithmic equations solvable! In the next section, we'll tackle the resulting exponential equation and find the value of 'x'.

Solving the Exponential Equation

Alright, we've successfully transformed our logarithmic equation into the exponential form: 2¹⁰ = x + 3. Now, it's time to roll up our sleeves and solve for 'x'. The first step here is to calculate 2¹⁰. If you have a calculator handy, this is a breeze. If not, remember that 2¹⁰ means 2 multiplied by itself 10 times (2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2). Either way, you'll find that 2¹⁰ equals 1024. So, our equation now looks like this: 1024 = x + 3. We're almost there! To isolate 'x', we need to get rid of the '+ 3' on the right side of the equation. The way we do this is by performing the inverse operation: subtraction. We subtract 3 from both sides of the equation to maintain the balance. This gives us 1024 - 3 = x + 3 - 3, which simplifies to 1021 = x. Voila! We've found the value of 'x'. It's like solving a puzzle, where each step brings us closer to the final answer. Solving this exponential equation involved a simple calculation and a basic algebraic manipulation. Now that we have a potential solution, it's important to take one more crucial step to ensure our answer is valid. We'll discuss this verification process in the next section.

Verifying the Solution

We've arrived at a potential solution: x = 1021. However, in the world of logarithmic equations, it's crucial to verify our solution. This is because logarithms have domain restrictions. Specifically, the argument of a logarithm (the expression inside the logarithm) must be greater than zero. If we plug in a value for 'x' that makes the argument zero or negative, the logarithm is undefined, and our solution is invalid. So, let's take our solution, x = 1021, and plug it back into the original equation: log₂(x+3) = 10. Substituting x = 1021, we get log₂(1021 + 3) = log₂(1024). Now, we need to check if log₂(1024) actually equals 10. We know that 2¹⁰ = 1024, so log₂(1024) indeed equals 10. Our solution checks out! This verification step is like the final seal of approval, confirming that our hard work has paid off. It ensures that our solution is not only mathematically correct but also makes sense in the context of the original logarithmic equation. Skipping this step can lead to incorrect answers, so always remember to verify your solutions when dealing with logarithmic equations. In our case, we can confidently say that x = 1021 is the correct solution. Let's wrap up our discussion with a final summary of the steps we took.

Conclusion

Awesome! We've successfully solved the logarithmic equation log₂(x+3) = 10. Let's quickly recap the steps we took to get there:

  1. Understood Logarithms: We started by making sure we understood the fundamental concept of logarithms as the inverse of exponentiation.
  2. Converted to Exponential Form: We converted the logarithmic equation into its equivalent exponential form: 2¹⁰ = x + 3. This step was crucial for eliminating the logarithm.
  3. Solved the Exponential Equation: We calculated 2¹⁰ and then solved for 'x' by subtracting 3 from both sides, resulting in x = 1021.
  4. Verified the Solution: We plugged our solution, x = 1021, back into the original equation to ensure it was valid and didn't result in taking the logarithm of a non-positive number.

By following these steps, you can confidently solve a wide range of logarithmic equations. Remember, the key is to understand the relationship between logarithms and exponents and to carefully apply the rules of algebra. Keep practicing, and you'll become a pro at solving these types of problems! If you have any more questions or want to try another example, feel free to ask. Happy solving, guys!