Simplifying Expressions & Logarithms: Math Problems Solved

Hey guys! Today, we're diving into a couple of math problems that involve simplifying expressions and dealing with logarithms. These types of questions are super common in math classes, so let's break them down step by step to make sure we understand exactly how to solve them. Get ready to sharpen those pencils and get your thinking caps on! We're gonna make math less intimidating and a whole lot more fun. Let's jump right in!

Simplifying Algebraic Expressions

Okay, let's tackle the first problem: finding the simplified form of the algebraic expression. This is where we really get to flex our algebraic muscles! The expression we're working with is $\frac{a^5x^2y^46^1}{83xy^1}$. Now, when you first look at something like this, it might seem a bit overwhelming, but trust me, it's totally manageable. The key here is to break it down into smaller, more digestible parts. We're going to use the basic rules of exponents and fractions to simplify this beast. Think of it like this: we're detectives, and each part of the expression is a clue. We just need to put the clues together in the right way.

First up, let's focus on the constants. We have $a^5$ in the numerator and 83 in the denominator. Since there are no common factors here, we'll just leave them as they are for now. Now, let’s zoom in on the $x$ terms. We have $x^2$ on top and $x$ on the bottom. Remember the rule of exponents that says when you're dividing like bases, you subtract the exponents? So, $x^2$ divided by $x$ is just $x^{2-1}$, which simplifies to $x$. See? We're already making progress! Next in line are the $y$ terms. We've got $y^4$ in the numerator and $y^1$ (or simply $y$) in the denominator. Applying the same exponent rule, $y^4$ divided by $y$ becomes $y^{4-1}$, which is $y^3$. We're slicing and dicing this expression like pros! Finally, we have the term $6^1$ in the numerator. This is just 6, so we'll keep it as is for the moment. Now, let's put all the pieces back together. We've got our constants, our simplified $x$ term, our simplified $y$ term, and our remaining constant. Combining them all, we get $\frac{a^5 * 6 * x * y^3}{83}$. It's much cleaner already, right? But hold on, can we simplify it even further? Let's take another look. Are there any common factors between the numerator and the denominator? In this case, it doesn't seem like it. So, this might just be our final simplified form. It’s always a good idea to double-check your work to make sure you haven’t missed anything, though. Go back through each step and ensure you've applied the exponent rules and fraction simplifications correctly. You might even want to try plugging in some numbers for the variables to see if both the original expression and the simplified one give you the same result. This is a neat trick for verifying your work. Remember, practice makes perfect when it comes to simplifying algebraic expressions. The more you do it, the more comfortable you'll become with the rules and the quicker you'll be able to spot opportunities for simplification. So, don't be afraid to tackle those complex-looking expressions – break them down, apply the rules, and you'll be simplifying like a math whiz in no time!

Evaluating Logarithmic Expressions

Alright, let’s shift gears and dive into the second part of our math adventure: evaluating logarithmic expressions. Logarithms might sound a bit intimidating at first, but they're really just another way of thinking about exponents. Trust me, once you get the hang of it, they're not as scary as they seem! The expression we're going to tackle is $^2\text{log } 81 - ^2\text{log } 27 + ^2\text{log } 9$. Notice that all these logarithms have the same base: 2. That's a crucial detail, because it means we can use some handy logarithm properties to simplify things. The first property we're going to use is the one that says when you subtract logarithms with the same base, you can combine them into a single logarithm by dividing the arguments. In other words, $^b\text{log } x - ^b\text{log } y = ^b\text{log } (\frac{x}{y})$. So, let's apply that to our first two terms: $^2\text{log } 81 - ^2\text{log } 27$. We can rewrite this as $^2\text{log } (\frac{81}{27})$. Now, what's 81 divided by 27? It's 3! So, our expression simplifies to $^2\text{log } 3$. Awesome! But we're not done yet. We still have that $+ ^2\text{log } 9$ hanging out. So, our expression now looks like $^2\text{log } 3 + ^2\text{log } 9$. Time for another logarithm property! This one says that when you add logarithms with the same base, you can combine them into a single logarithm by multiplying the arguments. So, $^b\text{log } x + ^b\text{log } y = ^b\text{log } (x * y)$. Applying this to our expression, we get $^2\text{log } (3 * 9)$, which simplifies to $^2\text{log } 27$. We're almost there! Now, we need to figure out what $^2\text{log } 27$ actually equals. Remember, a logarithm is just asking the question: