Solving Math Problems With Exponents And Algebra
Hey everyone! Today, we're diving into some exciting math problems involving exponents and algebraic expressions. We'll break down each step, so don't worry if it looks a bit intimidating at first. We'll tackle these problems together, making sure everyone understands the concepts. So, grab your pencils and let's get started!
If a = 5, b = ¼, and C = 2, Determine the Following Results
Let's jump right into our first challenge! We're given the values of three variables: a = 5, b = ¼, and C = 2. Our mission is to find the values of two expressions using these variables. This is where our knowledge of exponents and order of operations (PEMDAS/BODMAS) will come in handy. Remember, PEMDAS/BODMAS tells us the order in which we should perform operations: Parentheses/Brackets, Exponents/Orders, Multiplication and Division (from left to right), and Addition and Subtraction (from left to right).
a. ab²c⁵
In this part, our mission is to find the value of the expression ab²c⁵. This expression involves multiplication and exponents. Remember, the exponent applies only to the term it's directly attached to. So, b² means b multiplied by itself, and c⁵ means c multiplied by itself five times. To solve this, we'll substitute the given values of a, b, and c into the expression and then follow the order of operations.
First, let's talk about substitution. Substitution is simply replacing a variable with its given value. In this case, we'll replace 'a' with 5, 'b' with ¼, and 'c' with 2. This gives us: 5 * (¼)² * 2⁵. Now, we need to deal with the exponents. Remember that (¼)² means (¼) * (¼), which equals 1/16. Also, 2⁵ means 2 * 2 * 2 * 2 * 2, which equals 32. So, our expression now looks like this: 5 * (1/16) * 32. Next up is multiplication. We can multiply these numbers together from left to right. 5 multiplied by 1/16 equals 5/16. Then, we multiply 5/16 by 32. To do this, we can think of 32 as 32/1. So, we have (5/16) * (32/1). Multiplying the numerators gives us 5 * 32 = 160, and multiplying the denominators gives us 16 * 1 = 16. Therefore, we have 160/16. Lastly, we simplify the fraction 160/16. Both 160 and 16 are divisible by 16. 160 divided by 16 is 10, and 16 divided by 16 is 1. So, 160/16 simplifies to 10/1, which is simply 10. So, the final answer for ab²c⁵ is 10.
b. a³b⁻²c²/a²b⁻¹c⁴
Now, let's tackle the second expression: a³b⁻²c²/a²b⁻¹c⁴. This looks a bit more complex, but don't worry, we'll break it down. The key here is to remember the rules of exponents, especially when dealing with division and negative exponents. The rule for dividing exponents with the same base says that when you divide powers with the same base, you subtract the exponents. For example, x⁵ / x² = x^(5-2) = x³. Also, remember that a negative exponent means we take the reciprocal of the base raised to the positive exponent. For instance, x⁻² = 1/x². Now, let's apply these rules to our expression.
We start by separating the terms with the same base: (a³/a²) * (b⁻²/b⁻¹) * (c²/c⁴). Now, we apply the rule for dividing exponents with the same base. For the 'a' terms, we have a³ / a² = a^(3-2) = a¹. For the 'b' terms, we have b⁻² / b⁻¹ = b^(-2 - (-1)) = b^(-2 + 1) = b⁻¹. And for the 'c' terms, we have c² / c⁴ = c^(2-4) = c⁻². So, our expression now looks like this: a¹ * b⁻¹ * c⁻². Next, let's deal with the negative exponents. We know that b⁻¹ = 1/b and c⁻² = 1/c². So, we can rewrite our expression as a¹ * (1/b) * (1/c²). This can also be written as a / (bc²). Finally, we substitute the values of a, b, and c: a = 5, b = ¼, and c = 2. This gives us 5 / ((¼) * 2²). Let's simplify the denominator first. 2² equals 4, so we have 5 / ((¼) * 4). (¼) * 4 equals 1, so our expression simplifies to 5 / 1, which is simply 5. Therefore, the final answer for a³b⁻²c²/a²b⁻¹c⁴ is 5.
Simplify in the Form x^m y^n!
Alright, let's move on to the next type of problem, where we need to simplify expressions into the form x^m y^n. This means we want to combine all the 'x' terms and all the 'y' terms into single terms with exponents. We'll use the rules of exponents we discussed earlier, such as the product of powers rule (x^m * x^n = x^(m+n)), the quotient of powers rule (x^m / x^n = x^(m-n)), and the power of a power rule ((xm)n = x^(m*n)). Remember, the goal is to have only one 'x' term and one 'y' term, each with its own exponent.
a. (x³y⁶/x⁵y⁻²)^5
This expression involves a fraction raised to a power. Our strategy will be to simplify the fraction inside the parentheses first, and then apply the outer exponent. To simplify the fraction, we'll use the quotient of powers rule. Let's rewrite the expression as ((x³ / x⁵) * (y⁶ / y⁻²))^5. Now, let's simplify the 'x' terms and the 'y' terms separately. For the 'x' terms, we have x³ / x⁵ = x^(3-5) = x⁻². For the 'y' terms, we have y⁶ / y⁻² = y^(6 - (-2)) = y^(6 + 2) = y⁸. So, our expression inside the parentheses simplifies to x⁻²y⁸. Now, we have (x⁻²y⁸)⁵. Next, we apply the power of a product rule, which states that (xy)^n = x^n * y^n. This means we need to raise both x⁻² and y⁸ to the power of 5. We also use the power of a power rule, which tells us that (xm)n = x^(mn). So, (x⁻²)⁵ = x^(-25) = x⁻¹⁰, and (y⁸)⁵ = y^(8*5) = y⁴⁰. Therefore, our simplified expression is x⁻¹⁰y⁴⁰. We've successfully expressed it in the form x^m y^n!
b. (x⁻²y³)4/(x³y⁻¹)-³(x⁷y⁻²)^-1
This expression looks like a real challenge! But don't worry, we'll tackle it step by step. We have a fraction with expressions raised to powers in both the numerator and the denominator. Our approach will be to first simplify the numerator and denominator separately, and then simplify the resulting fraction. Let's start with the numerator: (x⁻²y³)^4. We use the power of a product rule and the power of a power rule. (x⁻²)^4 = x^(-24) = x⁻⁸, and (y³)^4 = y^(34) = y¹². So, the numerator simplifies to x⁻⁸y¹². Now, let's move on to the denominator: (x³y⁻¹)-³(x⁷y⁻²)-1. We need to simplify each term in the denominator separately first. For the first term, (x³y⁻¹)^-³, we use the power of a product rule and the power of a power rule. (x³)^-³ = x^(3*-3) = x⁻⁹, and (y⁻¹)^-³ = y^(-1*-3) = y³. So, the first term simplifies to x⁻⁹y³. For the second term, (x⁷y⁻²)^-1, we do the same thing. (x⁷)^-1 = x^(7*-1) = x⁻⁷, and (y⁻²)^-1 = y^(-2*-1) = y². So, the second term simplifies to x⁻⁷y². Now, we can rewrite the denominator as (x⁻⁹y³)(x⁻⁷y²). Next, we multiply these two terms together. To do this, we use the product of powers rule, which says that x^m * x^n = x^(m+n). So, x⁻⁹ * x⁻⁷ = x^(-9 + (-7)) = x⁻¹⁶, and y³ * y² = y^(3+2) = y⁵. Therefore, the denominator simplifies to x⁻¹⁶y⁵. Now, our entire expression looks like this: (x⁻⁸y¹²) / (x⁻¹⁶y⁵). Finally, we need to simplify this fraction. We use the quotient of powers rule, which says that x^m / x^n = x^(m-n). For the 'x' terms, we have x⁻⁸ / x⁻¹⁶ = x^(-8 - (-16)) = x^(-8 + 16) = x⁸. For the 'y' terms, we have y¹² / y⁵ = y^(12-5) = y⁷. So, our simplified expression is x⁸y⁷. We've successfully expressed it in the form x^m y^n! You guys are doing great!
Conclusion
And there we have it! We've successfully solved some challenging math problems involving exponents and algebraic expressions. We covered how to substitute values into expressions, how to use the rules of exponents to simplify expressions, and how to express expressions in the form x^m y^n. Remember, the key to mastering these concepts is practice, practice, practice! The more you work with these types of problems, the more comfortable you'll become with them. Keep up the great work, guys, and don't be afraid to ask questions. Math can be fun and rewarding, and with a little effort, you can conquer any challenge!
