Solving Logarithmic Equations: Finding The Value Of X

Hey guys, let's dive into a cool math problem! We're gonna tackle a logarithmic equation and figure out how to solve for x. The equation we're looking at is: $\log _2(7 x+6)=5$. This might look a little intimidating at first, but trust me, it's not as scary as it seems! We'll break it down step-by-step, making sure you understand every part of the process. Understanding logarithms is super important, especially if you're into algebra or any field that uses math. So, let's get started and uncover the secret to finding the value of x! This journey is all about understanding the relationship between logarithms and exponents and, ultimately, simplifying and solving equations. Remember that logarithms and exponents are inverse operations, just like addition and subtraction, or multiplication and division. The core idea is to rewrite the logarithmic equation in its equivalent exponential form. This allows us to apply our knowledge of exponents to isolate x and find its value. So, take a deep breath, get your thinking caps on, and let's unravel this mathematical puzzle together. We'll go through the fundamentals, which is essentially the transformation of logarithmic expressions to exponential equations, then proceed to the necessary algebraic manipulation, and, finally, we will verify our answer. This entire process is crafted to ensure a solid grasp of how to deal with logarithmic equations.

Understanding Logarithms

First things first, before we start solving, let's quickly review what a logarithm actually is. Basically, a logarithm answers the question: "To what power must we raise the base to get a certain number?" In our equation, $\log _2(7x + 6) = 5$, the base is 2. The logarithm (which is 5) tells us that 2 raised to the power of 5 equals $(7x + 6)$. This concept is super important for changing logarithmic equations into a format we can easily work with. Think of it like a secret code: logarithms are just a different way of writing exponents. To rephrase this, consider the general form of a logarithm: $\log_b(a) = c$. This means that $b^c = a$. Now, let's look at our specific equation, $\log _2(7x+6)=5$. Here, our b (base) is 2, our a is $(7x + 6)$, and our c (the result of the logarithm) is 5. So, according to the rule, we can rewrite this as $2^5 = 7x + 6$. See? It's all about switching between these two forms. Once we've got it in exponential form, solving for x becomes much easier because we can use our regular algebra skills. This conversion is the key to solving the problem. So, the essential takeaway is the ability to recognize and convert between logarithmic and exponential forms, allowing you to manipulate and solve the equations efficiently. Understanding this principle is crucial, not just for this problem, but for dealing with any logarithmic equation that you might encounter.

Converting to Exponential Form

Alright, now that we're all on the same page about what logarithms are, let's convert our equation from its logarithmic form to its exponential form. As we just discussed, the equation $\log _2(7x + 6) = 5$ translates to $2^5 = 7x + 6$. We've essentially rewritten the equation, making it much easier to solve. Notice how the base of the logarithm (which is 2) becomes the base of the exponent. The number on the other side of the equals sign (which is 5) becomes the exponent. And the expression inside the parentheses of the logarithm $(7x + 6)$ is equal to the result. This conversion step is absolutely crucial because it allows us to eliminate the logarithm and work with a standard algebraic equation. Now, let's take a look at the equation $2^5 = 7x + 6$. We know that $2^5$ means 2 multiplied by itself 5 times, which equals 32. So, we can simplify our equation to $32 = 7x + 6$. This is now a simple, straightforward linear equation, and solving it is within reach. By converting to exponential form, we've transformed the equation into something manageable, and the value of x is now accessible through simple algebraic methods. Remember, this step is all about making the equation easier to work with, and it sets the stage for isolating x in the following steps. This conversion is a fundamental skill that underpins solving logarithmic equations, giving you a powerful tool in your mathematical toolkit.

Solving for x

Okay, awesome, we've got our equation in exponential form: $32 = 7x + 6$. Now, let's solve for x! Our goal is to isolate x on one side of the equation. To do this, we'll use a few algebraic steps. First, we need to get rid of that +6. We can do this by subtracting 6 from both sides of the equation. This gives us $32 - 6 = 7x + 6 - 6$, which simplifies to $26 = 7x$. Now we have $26 = 7x$. Next, we need to get x all by itself. We do this by dividing both sides of the equation by 7. This gives us $\frac{26}{7} = \frac{7x}{7}$. When we simplify this, we get $x = \frac{26}{7}$. Congrats, guys! We've solved for x! So, the value of x that satisfies the original logarithmic equation is $\frac{26}{7}$. These steps involve basic algebraic principles that ensure we can isolate the variable of interest. The process includes subtraction and division, applied to both sides of the equation. Remember, when manipulating equations, it's vital to perform the same operations on both sides to maintain the equation's balance. This consistent approach leads to a reliable solution for x. The fundamental aim is to isolate x until it stands alone, which allows us to find its value. This is a crucial skill for not only solving this equation, but for handling many different equations in algebra.

Verification of Solution

Alright, we have found that $x = \frac{26}{7}$. It's always a super good idea to check our work to make sure we didn't make any mistakes along the way. We will plug our solution back into the original equation and see if it works. Our original equation was $\log _2(7x + 6) = 5$. So, we'll substitute $\frac{26}{7}$ for x: $\log _2(7 * (\frac{26}{7}) + 6) = 5$. First, let's simplify the term inside the logarithm: $7 * (\frac{26}{7})$ simplifies to 26, so our equation becomes $\log _2(26 + 6) = 5$. Now, simplify further: $26 + 6 = 32$. This gives us $\log _2(32) = 5$. So we ask ourselves, "2 to what power equals 32?" Well, $2^5 = 32$, which is what we expected. This is the same value on the right side of the equals sign in our original equation. Since both sides of the equation match up, we know our solution is correct. This verification process is a super important step. The key here is to make sure our solution does indeed make the original equation true. Substituting the solution back into the equation helps to ensure that no errors were made. The goal is to obtain a true statement, confirming the validity of our result. If the left-hand side equals the right-hand side after substitution, our result is correct. Verification is an integral part of problem-solving. It's a key practice that makes sure our answers are accurate, and our understanding is solid.

Conclusion

Woohoo! We did it, guys! We successfully solved the logarithmic equation $\log _2(7x + 6) = 5$. We went from a seemingly complex equation to finding the value of x, which is $\frac{26}{7}$. We began by understanding the fundamentals of logarithms and how to rewrite them as exponents. Then, we used that knowledge to convert the equation into exponential form, making it easier to solve. We used algebraic techniques to isolate x and find its value. Finally, we checked our answer to make sure it was correct. This whole process shows that even seemingly complex math problems are solvable if you break them down into smaller steps and understand the basic concepts. You've now gained valuable skills in solving logarithmic equations, which are applicable in various mathematical and scientific fields. Remember, practice is super important, so try more logarithmic equations to strengthen your skills. This is a very essential type of mathematical skill, and understanding how to solve these problems unlocks access to more advanced mathematical concepts. Keep practicing, and you'll become a pro at solving logarithmic equations in no time! Keep up the awesome work, and happy calculating, everyone!