Solve Logarithmic Equation: Exact Solutions

Hey math whizzes! Today, we're diving deep into the awesome world of logarithmic equations. These bad boys can seem a bit tricky at first, but trust me, once you get the hang of the rules, they're super fun to solve. Our mission today is to tackle this beast: $3 \log _5(5-3 z)+4=13$. We're not just looking for any old answer, guys; we need the exact solutions, which means no decimal approximations unless absolutely necessary. Get ready to flex those math muscles because we're about to break this down step-by-step!

First off, let's talk about what a logarithmic equation even is. Basically, it's an equation where the variable you're trying to solve for is inside a logarithm. Remember those magical little things called logarithms? They're the inverse of exponentiation. So, if you have $\log_b(x) = y$, it means $b^y = x$. This inverse relationship is super key to solving these problems. We're going to use this property to isolate the logarithm first, and then we can 'undo' the logarithm by converting it into an exponential form. It's like a puzzle, and each piece (or property) helps us get closer to the solution. So, when you see $\log _5$, that little '5' is our base. It tells us what number we need to raise something to in order to get another number. Keep that in mind as we move through the problem. We've got this!

Now, let's get down to business with our specific equation: $3 \log _5(5-3 z)+4=13$. Our first goal is to get that logarithm term, $\log _5(5-3 z)$, all by itself on one side of the equation. Think of it like peeling an onion, you gotta get to the core. We've got a few operations happening around the logarithm: multiplication by 3 and addition of 4. To isolate it, we'll reverse these operations using the order of operations in reverse (PEMDAS backwards, SADMEP if you will). So, the first thing we'll tackle is that '+4'. To get rid of it, we subtract 4 from both sides of the equation. This is a fundamental rule in algebra: whatever you do to one side, you must do to the other to keep the equation balanced. So, $3 \log _5(5-3 z)+4 - 4 = 13 - 4$. This simplifies nicely to $3 \log _5(5-3 z) = 9$. See? We're already one step closer to cracking this code! This simplification is crucial because it brings us nearer to isolating the logarithmic expression, which is the main hurdle we need to overcome before we can convert to exponential form.

Alright, we've got $3 \log _5(5-3 z) = 9$. Now, that logarithm is still being multiplied by 3. To isolate it completely, we need to undo that multiplication. The opposite of multiplying by 3 is dividing by 3, right? So, we divide both sides of the equation by 3. Again, gotta keep it balanced! This gives us $\frac{3 {\log _5(5-3 z)}}{3} = \frac{9}{3}$. Performing the division, we get $\log _5(5-3 z) = 3$. Boom! We've successfully isolated the logarithmic term. This is a major victory, guys! It means we've done the hard part of rearranging the equation to prepare it for the next, and arguably most exciting, step: converting it into an exponential equation. This isolation step is where many algebraic manipulations come into play, ensuring that we're setting ourselves up for a clean conversion. Remember, every step we take is deliberate and based on established mathematical principles, making the process logical and solvable.

Now that we have $\log _5(5-3 z) = 3$, it's time to unleash the power of the inverse relationship between logarithms and exponents. Remember our definition? If $\log_b(x) = y$, then $b^y = x$. In our case, the base $b$ is 5, the argument of the logarithm (what's inside the parentheses) is $5-3z$, and the value $y$ is 3. So, we can rewrite our logarithmic equation as an exponential one: $5^3 = 5-3z$. This conversion is the game-changer because it transforms the logarithmic equation, which can be tricky to solve directly for the variable, into a simple linear equation that we know how to handle. It's like switching from a foreign language to one you're fluent in. $5^3$ is straightforward to calculate: $5 \times 5 \times 5 = 125$. So, our equation becomes $125 = 5-3z$. This step is absolutely critical, as it bridges the gap between logarithmic and exponential forms, allowing us to proceed with standard algebraic techniques to find the value of $z$. Don't forget the base! It's the foundation upon which the exponential form is built.

We're in the home stretch now, folks! We have the linear equation $125 = 5-3z$. Our goal is to isolate $z$. We want to get all the terms with $z$ on one side and the constant terms on the other. Let's start by moving the constant term '5' from the right side to the left side. To do this, we subtract 5 from both sides: $125 - 5 = 5 - 5 - 3z$. This simplifies to $120 = -3z$. Now, $z$ is being multiplied by -3. To get $z$ by itself, we need to divide both sides by -3. So, $\frac{120}{-3} = \frac{-3z}{-3}$. Performing the division, we get $-40 = z$. And there you have it: $z = -40$. This is our potential solution! It's a clean, exact number, exactly what we were aiming for. This linear equation stage is where the algebraic manipulation really shines, turning a complex logarithmic problem into a simple solvable equation. Remember, every variable term and constant term needs careful attention as we rearrange the equation to isolate our target variable.

Before we celebrate, there's one super important step we absolutely cannot skip when solving logarithmic equations: checking our answer! Why? Because the argument of a logarithm (what's inside the parentheses) must always be positive. If we plug our solution back into the original equation and the argument turns out to be zero or negative, then that solution is extraneous and we have to throw it out. Let's check $z = -40$ in our original equation's argument: $5-3z$. Substitute $z=-40$: $5 - 3(-40) = 5 - (-120) = 5 + 120 = 125$. Since 125 is positive, our solution $z = -40$ is valid! This check is non-negotiable, guys. It ensures that we are not introducing invalid solutions through the process of converting to exponential form. The domain of a logarithmic function is strictly positive arguments, so verifying this is as crucial as the algebraic steps themselves. It's the final quality control for our math! So, the solution set is $\{-40\}$.

So, to recap the journey, we started with $3 {\log _5(5-3 z)+4=13}$. We painstakingly isolated the logarithmic term by subtracting 4 and then dividing by 3, arriving at $\log _5(5-3 z) = 3$. Then, we transformed this into an exponential equation, $5^3 = 5-3z$, which simplified to $125 = 5-3z$. Through basic algebra, we solved for $z$, getting $z = -40$. Finally, and most critically, we verified our solution by plugging it back into the argument of the logarithm to ensure it was positive. Since $5-3(-40) = 125$, which is indeed positive, our solution $z = -40$ is confirmed. The solution set for this logarithmic equation is $\{-40\}$. Keep practicing these steps, and you'll become a log equation pro in no time! Remember, consistency and attention to detail are your best friends in mathematics. Each step builds upon the last, and a thorough understanding of each principle allows you to confidently tackle even more complex problems. Happy solving!