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Chapter 0: Problem 97

Simplify using properties of exponents. $$\left(25 x^{4} y^{6}\right)^{\frac{1}{2}}$$

Short Answer

Expert verified
The simplified version of \( (25 x^{4} y^{6})^{1/2} \) is \( 5x^{2}y^{3}\).

Step by step solution

01

Power of a Product Rule

First, apply the power of a product rule. This rule states that \((ab)^n = a^n b^n\). So, we have: \((25 x^{4} y^{6})^{1/2} = 25^{1/2} * (x^{4})^{1/2} * (y^{6})^{1/2}\).
02

Power of a Power Rule

Now, apply the power of a power rule which states that \((a^n)^m = a^{n*m}\). So, we have: \(25^{1/2} * (x^{4})^{1/2} * (y^{6})^{1/2} = 5 * x^{4/2} * y^{6/2}\).
03

Simplify the Expression

Finally, simplify the expression by simplifying the exponents. The expression becomes: \(5 * x^{2} * y^{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Power of a Product Rule: Simplifying Multiple Terms
When you encounter an expression like \((25 x^{4} y^{6})^{\frac{1}{2}}\), the power of a product rule is your go-to tool. This rule says that if you have a product raised to a power like
  • \((ab)^n\)
It can be separated into
  • \(a^n \times b^n\).
This means every element inside the bracket gets individually raised to the power outside. So,
  • \((25 x^{4} y^{6})^\frac{1}{2}\) becomes \(25^{\frac{1}{2}} \times (x^{4})^{\frac{1}{2}} \times (y^{6})^{\frac{1}{2}}\).
Breaking it down like this makes it easier to handle each component separately, especially when you move on to the next rules or steps in simplification.
Remember to apply the exponent to each term inside the bracket!
Power of a Power Rule: Managing Exponentiated Exponents
The power of a power rule is handy when dealing with terms that already have an exponent, and you are raising it to yet another power. According to this rule:
  • \((a^n)^m = a^{n \times m}\).
Let's apply it to our individual components from the previous section:
  • \(25^{\frac{1}{2}}\),
  • \((x^4)^\frac{1}{2} = x^{4 \times \frac{1}{2}} = x^2\),
  • \((y^6)^\frac{1}{2} = y^{6 \times \frac{1}{2}} = y^3\).
This helps to simplify complex expressions involving multiple layers of exponents by reducing them to a single layer. The rule essentially allows you to "multiply" the exponents when the base remains the same. This simplification step is crucial in ensuring the expression is more manageable and comprehensible.
Simplification of Expressions: Piecing It All Together
After using the power rules, the last step is to simplify the expression as a whole. This involves evaluating and combining what we obtained from previous steps. From our example, it becomes:
  • \(25^{\frac{1}{2}}\) simplifies to \(5\), since the square root of 25 is 5.
  • We already simplified \(x^{2}\) and \(y^{3}\).
Putting it all together:
  • \(5 \times x^2 \times y^3\).
Thus, the final simplified expression is \(5x^2y^3\). When simplifying, always be sure to check each term independently and see how they come together to form a neat and concise expression. The goal of simplification is to make the expression as straightforward as possible, making it easier to interpret and apply in any further calculations or problems.

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Most popular questions from this chapter

Will help you prepare for the material covered in the first section of the next chapter. If \(y=4-x^{2},\) find the value of \(y\) that corresponds to values of \(x\) for each integer starting with \(-3\) and ending with 3

Explain why \(|x|<-4\) has no solution.

Determine whether each statement makes sense or does not make sense, and explain your reasoning. Although I can solve \(3 x+\frac{1}{5}=\frac{1}{4}\) by first subtracting \(\frac{1}{5}\) from both sides, I find it easier to begin by multiplying both sides by \(20,\) the least common denominator.

If you are given a quadratic equation, how do you determine which method to use to solve it?

Will help you prepare for the material covered in the next section. Jane's salary exceeds Jim's by 150 dollar per week. If \(x\) represents Jim's weekly salary, write an algebraic expression that models Jane's weekly salary.
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