Math Simplification & Logarithm Problems Solved

Let's dive into these math problems together! We'll break down each one step-by-step so you can easily understand the solutions. We'll be covering simplification of algebraic expressions, radicals, and logarithms. Get ready to sharpen your math skills, guys!

1. Simplifying Algebraic Expressions with Exponents

Okay, so the first question asks us to find the simplified form of this expression:

((x^(-2)y^(3)z^(-1))/(x^(-1)y^(-2)z^(3)))^(-2)

This looks a bit intimidating, but don't worry, we'll tackle it together! The key here is to remember the rules of exponents. Let's break it down:

First, let's focus on simplifying the expression inside the parentheses. We have a fraction with variables raised to different powers. Remember these exponent rules:

• x^(-n) = 1/(x^n)
• (x^*m*)/(*x*n) = x^(m-n)

Let's apply these rules. We'll start by dealing with the x terms. We have x^(-2) in the numerator and x^(-1) in the denominator. Using the rule for division, we get:

x^(-2) / x^(-1) = x^(-2 - (-1)) = x^(-2 + 1) = x^(-1)

Next, let's look at the y terms. We have y^(3) in the numerator and y^(-2) in the denominator:

y^(3) / y^(-2) = y^(3 - (-2)) = y^(3 + 2) = y^(5)

Now, let's handle the z terms. We have z^(-1) in the numerator and z^(3) in the denominator:

z^(-1) / z^(3) = z^(-1 - 3) = z^(-4)

So, inside the parentheses, we now have:

x^(-1) y^(5) z^(-4)

But we're not done yet! We still have that exponent of -2 outside the parentheses. Remember this rule: (x^*m*)n = x^(m * n)

We need to apply this rule to each term inside the parentheses:

(x^(-1))(-2) = x^(-1 * -2) = x^(2)

(y⁽⁵⁾⁾(-2) = y^(5 * -2) = y^(-10)

(z^(-4))(-2) = z^(-4 * -2) = z^(8)

Putting it all together, we get:

x^(2) y^(-10) z^(8)

To get rid of the negative exponent, we can rewrite y^(-10) as 1/y^(10). So, the final simplified form is:

(x^(2) * z^(8)) / y^(10)

Therefore, the simplified form of the expression is (x^(2) * z^(8)) / y^(10). Remember the exponent rules, guys, they're your best friends in these types of problems! This problem showcases how understanding and applying exponent rules systematically can lead to a simplified solution. The initial complex-looking expression is reduced to a much cleaner form by meticulously following the order of operations and exponent properties. Recognizing the power of negative exponents and how they translate to reciprocals is crucial in arriving at the final answer. Moreover, the ability to break down a large problem into smaller, manageable steps is a valuable skill in mathematics and problem-solving in general. By focusing on each variable separately and then combining the results, we avoid getting overwhelmed by the complexity of the entire expression. This approach highlights the importance of methodical thinking and attention to detail when dealing with mathematical challenges. The final simplified expression not only provides the answer but also offers a deeper understanding of the relationship between the variables and their exponents, reinforcing the fundamental concepts of algebra. Thus, mastering exponent rules is essential for anyone pursuing further studies in mathematics and related fields.

2. Simplifying Radicals

Next up, we need to simplify this expression involving radicals:

(2√6)/(√3+√2)

The key to simplifying expressions like this is to rationalize the denominator. This means getting rid of the radicals in the denominator. We do this by multiplying both the numerator and the denominator by the conjugate of the denominator.

The conjugate of √3 + √2 is √3 - √2. Remember, the conjugate is formed by simply changing the sign between the terms. Let's multiply both the numerator and denominator by this conjugate:

[(2√6) / (√3 + √2)] * [(√3 - √2) / (√3 - √2)]

Now, let's multiply out the numerator:

2√6 * (√3 - √2) = 2√6 * √3 - 2√6 * √2 = 2√(63) - 2√(62) = 2√18 - 2√12

We can simplify these radicals further. Remember to look for perfect square factors:

• √18 = √(9 * 2) = √9 * √2 = 3√2
• √12 = √(4 * 3) = √4 * √3 = 2√3

So, the numerator becomes:

2 * 3√2 - 2 * 2√3 = 6√2 - 4√3

Now, let's multiply out the denominator. This is where the conjugate comes in handy! When you multiply conjugates, the middle terms cancel out:

(√3 + √2)(√3 - √2) = (√3 * √3) - (√3 * √2) + (√2 * √3) - (√2 * √2) = 3 - √6 + √6 - 2 = 3 - 2 = 1

Wow, the denominator simplified to just 1! That makes things much easier. So, our expression now is:

(6√2 - 4√3) / 1 = 6√2 - 4√3

Therefore, the simplified form of (2√6)/(√3+√2) is 6√2 - 4√3. Rationalizing the denominator is a powerful technique, guys! It allows us to eliminate radicals from the denominator and often leads to a much simpler expression. In this case, multiplying by the conjugate transformed the complex fraction into a straightforward difference of radicals. This technique is widely used in algebra and calculus, particularly when dealing with limits and derivatives involving radical expressions. The underlying principle is to exploit the difference of squares identity, which states that (a + b)(a - b) = a^2 - b^2. By choosing the conjugate as the multiplier, we effectively square both terms in the denominator, thereby removing the square roots. The process of simplifying radicals often involves identifying perfect square factors within the radicand (the number under the square root sign). Recognizing these factors allows us to rewrite the radical as a product of simpler radicals, ultimately leading to a reduced form. In this problem, we simplified √18 and √12 by extracting the perfect square factors 9 and 4, respectively. This step is crucial for obtaining the final simplified expression. Furthermore, the ability to combine like terms is essential for achieving the most concise form of the answer. In the numerator, we combined the terms involving √2 and √3 to arrive at the final expression 6√2 - 4√3. Mastering these techniques not only enhances algebraic skills but also fosters a deeper understanding of the properties of radicals and their manipulation.

3. Logarithm Problem

Alright, let's tackle this logarithm problem. We're given:

log 2 = a

log 3 = b

And we need to find log 360.

The key here is to express 360 as a product of its prime factors. Let's do that:

360 = 2 * 180 = 2 * 2 * 90 = 2 * 2 * 2 * 45 = 2 * 2 * 2 * 3 * 15 = 2 * 2 * 2 * 3 * 3 * 5 = 2^(3) * 3^(2) * 5

So, 360 = 2^(3) * 3^(2) * 5. Now we can use the logarithm product rule: log (x * y) = log x + log y

log 360 = log (2^(3) * 3^(2) * 5) = log (2^(3)) + log (3^(2)) + log 5

We also need to use the power rule: log (x^n) = n * log x

log (2^(3)) + log (3^(2)) + log 5 = 3 log 2 + 2 log 3 + log 5

We know log 2 = a and log 3 = b, so we can substitute those in:

3 log 2 + 2 log 3 + log 5 = 3a + 2b + log 5

Now, we need to figure out log 5. We can use the fact that log 10 = 1 and the quotient rule: log (x/y) = log x - log y

Since 5 = 10/2, we can write:

log 5 = log (10/2) = log 10 - log 2 = 1 - log 2 = 1 - a

Now we can substitute this back into our expression:

3a + 2b + log 5 = 3a + 2b + (1 - a) = 3a + 2b + 1 - a = 2a + 2b + 1

Therefore, log 360 = 2a + 2b + 1. This logarithm problem demonstrates the power of prime factorization and the fundamental properties of logarithms, including the product rule, power rule, and quotient rule, guys. By breaking down 360 into its prime factors, we could express the logarithm of 360 in terms of the given logarithms of 2 and 3. The ability to manipulate logarithmic expressions using these rules is essential for solving a wide range of problems in mathematics, science, and engineering. The key step in this solution is recognizing the relationship between log 5, log 10, and log 2. Since 5 is equal to 10 divided by 2, we can use the quotient rule to express log 5 as the difference between log 10 and log 2. Knowing that log 10 is equal to 1 (in base 10), we can then substitute the given value of log 2 to find the value of log 5 in terms of a. This clever manipulation allows us to express the final answer solely in terms of a and b, as required. Furthermore, this problem highlights the importance of understanding the base of the logarithm. While the base is not explicitly stated, it is assumed to be base 10, which is the common logarithm. If the base were different, the value of log 10 would also be different, affecting the final result. Thus, paying close attention to the base of the logarithm is crucial for accurate calculations. Mastering these techniques not only enhances problem-solving skills but also provides a deeper appreciation for the elegance and power of logarithmic functions.

4. Incomplete Question

The fourth question seems to be incomplete. Please provide the full question so I can help you solve it!

I hope this breakdown was helpful, guys! Remember to practice these techniques to become even more confident in your math skills.