Logarithmic Equation Solution: Log_2(3x-7)=3

Hey guys! Ever get stuck on a math problem and wish there was a simple, step-by-step guide to figure it out? Well, you're in luck because today we're diving deep into solving logarithmic equations. Specifically, we're tackling the beast: $\log _2(3 x-7)=3$. We'll break down exactly how to find the solution, and by the end of this, you'll be a log equation pro. Forget those confusing textbooks; we're making math make sense, Plastik Magazine style!

The Heart of the Matter: Understanding Logarithmic Equations

Alright, let's get down to brass tacks. What is a logarithmic equation like $\log _2(3 x-7)=3$? Think of it as the inverse of an exponential equation. The logarithm tells you what exponent you need to raise a specific base to, in order to get a certain number. In our case, the base is 2. The expression inside the logarithm, $(3x-7)$, is the number we're trying to reach, and the result, 3, is the exponent we're looking for. So, the equation $\log _2(3 x-7)=3$ is essentially asking: "To what power must we raise 2 to get the value of $(3x-7)$?" The answer, according to the equation, is 3. This fundamental understanding is key to unraveling the mystery of these equations. We need to isolate the variable, 'x', and the properties of logarithms are our best friends here. Remember, the goal is always to get 'x' by itself on one side of the equation. The techniques we use are derived from the very definition of logarithms, which states that if $\log_b(y) = x$, then $b^x = y$. This conversion is the golden ticket to transforming a logarithmic equation into a more manageable exponential one. So, when you see $\log _2(3 x-7)=3$, immediately think about that conversion. What's the base? It's 2. What's the exponent? It's 3. What's the result of that exponentiation? It's the expression inside the log, $(3x-7)$. This direct translation is the most crucial step in solving any logarithmic equation of this form. Mastering this conversion means you're already halfway there. It's like having the secret code to unlock the problem. Don't let the symbols intimidate you; they represent a clear relationship, and once you grasp that, the rest is just algebraic maneuvering. We'll explore how this conversion directly leads us to the solution we need.

Step-by-Step: Solving $\log _2(3 x-7)=3$ for 'x'

Now, let's put that understanding into action! To solve $\log _2(3 x-7)=3$, we're going to use the definition of a logarithm. As we just discussed, if $\log_b(y) = x$, then $b^x = y$. Applying this to our equation, we have:

• Base (b): 2
• Exponent (x): 3
• Result (y): $(3x-7)$

So, we can rewrite our logarithmic equation as an exponential one:

$$2^3 = 3x - 7$$

Easy peasy, right? We've just transformed a logarithmic equation into a simple linear one. Now, all we need to do is solve for 'x'.

First, calculate $2^3$:

$2^3 = 2 imes 2 imes 2 = 8$

So, our equation becomes:

$$8 = 3x - 7$$

Next, we want to isolate the term with 'x'. To do this, add 7 to both sides of the equation:

$$8 + 7 = 3x - 7 + 7$$

$$15 = 3x$$

Finally, to get 'x' all by itself, divide both sides by 3:

$$\frac{15}{3} = \frac{3x}{3}$$

$$5 = x$$

And there you have it! The solution to the equation $\log _2(3 x-7)=3$ is x = 5. It’s that straightforward when you break it down. This methodical approach ensures that we don't miss any steps and arrive at the correct answer. Remember to always check your work, especially when dealing with logarithms, to ensure the argument of the logarithm is positive. In this case, if $x=5$, then $3x-7 = 3(5)-7 = 15-7 = 8$, which is positive, so our solution is valid.

Checking Our Work: Is x=5 Really the Answer?

It's always a good idea, especially in math, to double-check your answers. Did we really find the correct value for 'x'? Let's plug x = 5 back into the original equation: $\log _2(3 x-7)=3$.

Substitute 5 for x:

$$\log _2(3(5) - 7)$$

Perform the multiplication and subtraction inside the parentheses:

$$\log _2(15 - 7)$$

$$\log _2(8)$$

Now, we ask ourselves: "What power do we need to raise 2 to, in order to get 8?" Well, we know that $2^1 = 2$, $2^2 = 4$, and $2^3 = 8$. So, the power is 3!

$$\log _2(8) = 3$$

This matches the right side of our original equation. Success! Our solution $x=5$ is absolutely correct. This verification step is crucial because it confirms that our algebraic manipulations were sound and that we haven't made any errors along the way. It also ensures that the argument of the logarithm ($3x-7$) remains positive, which is a requirement for logarithms to be defined in the real number system. If we had obtained a value for 'x' that made $3x-7$ zero or negative, that solution would be extraneous and invalid. In this instance, $3(5) - 7 = 15 - 7 = 8$, which is greater than zero, confirming the validity of our answer. This rigorous checking process builds confidence in our mathematical abilities and ensures accuracy.

Analyzing the Options: Why Other Choices Don't Cut It

So, we found our answer to be $x=5$. But what about those other options provided? Let's quickly see why they don't work, which can be a super helpful way to reinforce your understanding and catch potential mistakes. The options were A. $\frac{1}{3}$, B. 4, C. 5, D. $\frac{16}{3}$. We already confirmed that C. 5 is the correct answer.

Let's test Option A: $x = \frac{1}{3}$.

Plug $\frac{1}{3}$ into the original equation: $\log _2(3(\frac{1}{3}) - 7)$.

This simplifies to $\log _2(1 - 7)$, which is $\log _2(-6)$. Uh oh! We can't take the logarithm of a negative number in the real number system. So, $\frac{1}{3}$ is definitely not our solution.

Now, let's look at Option B: $x = 4$.

Substitute 4 into the equation: $\log _2(3(4) - 7)$.

This gives us $\log _2(12 - 7)$, which equals $\log _2(5)$. Is $\log _2(5)$ equal to 3? No, because $2^3 = 8$, not 5. So, 4 is incorrect.

Finally, let's check Option D: $x = \frac{16}{3}$.

Substitute $\frac{16}{3}$ into the equation: $\log _2(3(\frac{16}{3}) - 7)$.

This simplifies to $\log _2(16 - 7)$, which is $\log _2(9)$. Is $\log _2(9)$ equal to 3? Nope. Since $2^3 = 8$, and 9 is close but not equal to 8, $\frac{16}{3}$ is not the solution.

See? By systematically testing the incorrect options, we further validate that x = 5 is indeed the only correct answer. It's a great way to build confidence in your final result and ensure you've truly mastered the problem. This process is not just about finding the right answer, but understanding why it's right and why the others are wrong.

Beyond the Basics: When Logarithm Problems Get Trickier

While solving $\log _2(3 x-7)=3$ was pretty straightforward, logarithmic equations can get more complex, guys. Sometimes you'll encounter equations with logarithms on both sides, or equations that require using logarithm properties like the product rule, quotient rule, or power rule to simplify them before you can convert them to exponential form. For instance, an equation might look something like $\log(x) + \log(x-3) = 1$. Here, you'd first use the product rule to combine the logs: $\log(x(x-3)) = 1$. Then, you'd convert it to exponential form (assuming base 10 here, as it's often implied when no base is written): $10^1 = x(x-3)$. This would lead to a quadratic equation, $10 = x^2 - 3x$, which you'd then solve for 'x'. Remember, after solving, you must check your answers by plugging them back into the original equation to ensure the arguments of all logarithms are positive. An extraneous solution can easily pop up when you're dealing with more complicated forms. Another common scenario involves different bases, which might require using the change-of-base formula. Understanding these advanced techniques will equip you to handle virtually any logarithmic equation thrown your way. The core principle, however, always remains the same: understand the definition, use the properties, convert to exponential form if necessary, solve for the variable, and always, always check your solutions. Keep practicing, and these complex problems will start feeling much more manageable. The journey through logarithms is a rewarding one, revealing the elegant interplay between exponents and their inverse operations. So, don't shy away from the challenge; embrace it!

Conclusion: Mastering Logarithms One Equation at a Time

So there you have it! We've successfully tackled the logarithmic equation $\log _2(3 x-7)=3$ and found the solution to be x = 5. We broke it down by converting the log equation into its exponential form, $2^3 = 3x-7$, which simplified our task considerably. We then used basic algebra to isolate 'x', arriving at our final answer. We also took the crucial step of checking our solution by plugging it back into the original equation, confirming its validity and ensuring the argument of the logarithm was positive. Finally, we examined why the other multiple-choice options were incorrect, solidifying our understanding. Mathematics is all about practice, and by working through problems like this, you build the skills and confidence needed to ace any challenge. Keep exploring, keep solving, and remember that even complex equations can be unraveled with the right approach. Happy calculating, everyone!

The correct answer is C. 5.