Solve Log₂(5x + 4) = 2: Step-by-Step Guide

Hey guys! Today, let's dive into the world of logarithmic equations and tackle a fun problem. We've got this equation: log₂(5x + 4) = 2, and our mission is to find the value of 'x'. Don't worry, it's not as intimidating as it looks! We'll break it down step by step, so you can follow along easily.

Understanding Logarithms

Before we jump into solving, let's quickly refresh what logarithms are all about. Think of a logarithm as the inverse operation of exponentiation. In simpler terms, it answers the question: "To what power must we raise the base to get a certain number?" For example, log₂(8) = 3 because 2 raised to the power of 3 equals 8 (2³ = 8). The base here is 2, and the number we're trying to get is 8. The logarithm, 3, is the exponent we need.

Now, understanding the core concept of logarithms is super crucial for solving equations like the one we have. Logarithms are essentially the inverse of exponential functions. So, if we have an expression like logₐ(b) = c, it's the same as saying aᶜ = b. Think of it as a roundabout way of expressing exponents. This relationship is the key to unlocking logarithmic equations. When you see a logarithm, try to visualize it as an exponent in disguise. This will make the process of solving equations much more intuitive and less daunting. For instance, in our case, log₂(5x + 4) = 2 tells us that 2 raised to the power of 2 is equal to 5x + 4. This is the fundamental connection that allows us to transform the logarithmic equation into a simple algebraic one, which we can then solve using familiar techniques. The beauty of logarithms lies in their ability to simplify complex calculations and relationships, making them indispensable tools in various fields, from mathematics and physics to computer science and finance. So, mastering this fundamental relationship is not just about solving this particular problem, but about building a solid foundation for more advanced concepts in the future. Remember, practice makes perfect, so the more you work with logarithms, the more comfortable you'll become with them.

Converting Logarithmic to Exponential Form

The trick to solving our equation is to convert it from logarithmic form to exponential form. Remember that logₐ(b) = c is equivalent to aᶜ = b. Applying this to our equation, log₂(5x + 4) = 2, we get:

2² = 5x + 4

See? We've transformed the logarithmic equation into a simple algebraic equation. The base of the logarithm (2) becomes the base of the exponent, the result of the logarithm (2) becomes the exponent, and the argument of the logarithm (5x + 4) becomes the result of the exponentiation. This conversion is the crucial step in solving logarithmic equations. It allows us to eliminate the logarithm and work with a more familiar algebraic structure. By understanding this transformation, you can tackle a wide range of logarithmic problems. The ability to switch between logarithmic and exponential forms is like having a secret weapon in your math arsenal. It's not just about memorizing a formula; it's about understanding the underlying relationship between these two mathematical concepts. Once you grasp this connection, you'll find that solving logarithmic equations becomes much more intuitive and less mechanical. This conversion process is a testament to the power of mathematical notation, which allows us to represent complex relationships in a concise and manageable way. So, embrace this transformation, practice it, and you'll be well on your way to mastering logarithms!

Solving the Algebraic Equation

Now we have a straightforward equation to solve:

2² = 5x + 4

First, simplify 2²:

4 = 5x + 4

Next, we want to isolate 'x'. Subtract 4 from both sides:

4 - 4 = 5x + 4 - 4

0 = 5x

Now, divide both sides by 5:

0 / 5 = 5x / 5

0 = x

So, we've found that x = 0! Solving this algebraic equation involves the fundamental principles of algebra: simplifying, isolating the variable, and performing inverse operations. Each step is designed to gradually peel away the layers surrounding the variable 'x' until we can determine its value. In this case, we first simplified the exponent, then used subtraction to move the constant term to one side, and finally, used division to isolate 'x'. These are the bread and butter techniques of algebraic manipulation, and mastering them is essential for tackling a wide range of mathematical problems. The beauty of algebra lies in its systematic approach to problem-solving. By following a set of rules and procedures, we can unravel even complex equations and arrive at a solution. This process is not just about finding the answer; it's about developing logical thinking and problem-solving skills that are applicable in many areas of life. So, when you approach an algebraic equation, remember to break it down into smaller, manageable steps, and apply the appropriate operations to isolate the variable. With practice and perseverance, you'll become a confident algebraic problem-solver!

Checking the Solution

It's always a good idea to check our solution to make sure it works in the original equation. Let's plug x = 0 back into log₂(5x + 4) = 2:

log₂(5(0) + 4) = log₂(4)

Since 2² = 4, log₂(4) = 2. Our solution checks out!

Checking the solution is a crucial step in solving any equation, but it's especially important when dealing with logarithmic equations. Logarithms have domain restrictions, meaning that the argument of the logarithm (the expression inside the parentheses) must be positive. If we plug in a value for 'x' that makes the argument negative or zero, the logarithm is undefined, and the solution is invalid. In our case, we found x = 0, and when we plug it back into the original equation, we get log₂(5(0) + 4) = log₂(4). Since 4 is positive, our solution is valid. However, if we had found a value of 'x' that resulted in a negative or zero argument, we would have had to discard that solution. This check is not just about verifying our calculations; it's about ensuring that our solution makes sense within the context of the original problem. It's a safeguard against extraneous solutions, which are values that satisfy the transformed equation but not the original one. So, always remember to check your solutions, especially when working with logarithms, radicals, or rational expressions. This simple step can save you from making costly mistakes and ensure that you have a correct and meaningful answer.

Final Answer

Therefore, if log₂(5x + 4) = 2, then x = 0.

So there you have it! We've successfully solved a logarithmic equation by converting it to exponential form, solving the resulting algebraic equation, and checking our solution. Remember, the key is to understand the relationship between logarithms and exponents and to apply the basic principles of algebra. Keep practicing, and you'll become a pro at solving these types of problems!

"logarithmic equation", "exponential form", "algebraic equation", "checking solutions", "logarithms"