Simplifying Algebraic Expressions Step-by-Step Solution

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Simplifying algebraic expressions is a fundamental skill in mathematics, particularly in algebra and calculus. This article will walk you through a detailed, step-by-step solution for simplifying the expression (16y8)14Γ—y3{\left(\frac{16}{y^8}\right)^{\frac{1}{4}} \times y^3}. Understanding the process not only helps in solving this specific problem but also enhances your ability to tackle similar algebraic challenges. We will break down each step, explain the underlying mathematical principles, and ensure clarity so that you can confidently approach such problems in the future. Let’s dive into the intricacies of exponents, radicals, and algebraic manipulations.

Understanding the Problem

Before we begin, let’s clearly understand the expression we need to simplify:

(16y8)14Γ—y3{ \left(\frac{16}{y^8}\right)^{\frac{1}{4}} \times y^3 }

This expression involves a fraction raised to a fractional exponent, multiplied by a power of y{ y }. To simplify it effectively, we'll use several key algebraic properties, including the power of a quotient, the power of a power, and the multiplication of exponents with the same base. It’s crucial to recognize that the exponent 14{ \frac{1}{4} } represents the fourth root, and we'll use this understanding to break down the expression.

Key Concepts and Properties

To successfully simplify this expression, we need to be familiar with the following concepts and properties:

  1. Power of a Quotient: (ab)n=anbn{\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}}
  2. Power of a Power: (am)n=amn{(a^m)^n = a^{mn}}
  3. Product of Powers: amΓ—an=am+n{a^m \times a^n = a^{m+n}}
  4. Fractional Exponents: a1n=an{a^{\frac{1}{n}} = \sqrt[n]{a}}

These properties allow us to manipulate exponents and simplify complex expressions. Grasping these concepts is vital for not just this problem, but for a wide array of algebraic simplifications.

Step-by-Step Solution

Now, let's proceed with the step-by-step simplification of the expression. We'll apply the properties mentioned above to break down the expression into manageable parts.

Step 1: Apply the Power of a Quotient

First, we apply the power of a quotient rule to the term inside the parentheses:

(16y8)14=1614(y8)14{ \left(\frac{16}{y^8}\right)^{\frac{1}{4}} = \frac{16^{\frac{1}{4}}}{(y^8)^{\frac{1}{4}}} }

This step separates the numerator and the denominator, each raised to the power of 14{ \frac{1}{4} }. This makes it easier to handle each part individually.

Step 2: Simplify the Numerator

Next, we simplify the numerator 1614{ 16^{\frac{1}{4}} }. Recall that a1n=an{ a^{\frac{1}{n}} = \sqrt[n]{a} }, so we have:

1614=164{ 16^{\frac{1}{4}} = \sqrt[4]{16} }

The fourth root of 16 is 2 because 24=16{ 2^4 = 16 }. Therefore:

1614=2{ 16^{\frac{1}{4}} = 2 }

This simplifies the numerator to a straightforward constant value, making the expression cleaner.

Step 3: Simplify the Denominator

Now, let's simplify the denominator (y8)14{ (y^8)^{\frac{1}{4}} }. We apply the power of a power rule, which states (am)n=amn{ (a^m)^n = a^{mn} }:

(y8)14=y8Γ—14=y2{ (y^8)^{\frac{1}{4}} = y^{8 \times \frac{1}{4}} = y^2 }

By multiplying the exponents, we reduce the denominator to a simpler form involving y{ y } raised to a power.

Step 4: Substitute Simplified Terms Back into the Expression

Now that we've simplified both the numerator and the denominator, we substitute these back into our original expression:

1614(y8)14=2y2{ \frac{16^{\frac{1}{4}}}{(y^8)^{\frac{1}{4}}} = \frac{2}{y^2} }

So, our original expression now looks like:

2y2Γ—y3{ \frac{2}{y^2} \times y^3 }

This simplification makes the next step more straightforward.

Step 5: Multiply by y3{ y^3 }

Next, we multiply the simplified fraction by y3{ y^3 }. This step involves multiplying terms with the same base, which requires adding their exponents:

2y2Γ—y3=2Γ—y3y2{ \frac{2}{y^2} \times y^3 = 2 \times \frac{y^3}{y^2} }

To simplify further, we divide y3{ y^3 } by y2{ y^2 }. This is equivalent to subtracting the exponents:

y3y2=y3βˆ’2=y1=y{ \frac{y^3}{y^2} = y^{3-2} = y^1 = y }

Step 6: Final Simplification

Finally, we substitute this back into our expression:

2Γ—y=2y{ 2 \times y = 2y }

Thus, the simplified expression is:

2y{ 2y }

However, this result does not match any of the provided options. Let's re-evaluate our steps to identify any potential errors.

Re-evaluating the Solution

Upon reviewing the steps, it appears there was a misinterpretation in the final steps. Let's correct it.

Step 5 (Corrected): Multiply by y3{ y^3 }

We have the expression:

2y2Γ—y3{ \frac{2}{y^2} \times y^3 }

This can be rewritten as:

2Γ—y3y2{ 2 \times \frac{y^3}{y^2} }

When dividing terms with the same base, we subtract the exponents:

y3y2=y3βˆ’2=y1=y{ \frac{y^3}{y^2} = y^{3-2} = y^1 = y }

So, the expression becomes:

2Γ—y=2y{ 2 \times y = 2y }

Step 6 (Corrected): Final Simplification

Thus, the simplified expression is:

2y{ 2y }

However, we made a mistake in the calculation. Let's correct it again.

We had:

2y2Γ—y3{ \frac{2}{y^2} \times y^3 }

Which should be calculated as:

2y3βˆ’2=2y1=2y{ 2y^{3-2} = 2y^1 = 2y }

It seems we still have an issue because this result 2y{ 2y } is not among the options provided (A) 2y5{ \frac{2}{y^5} }, (B) 4y5{ \frac{4}{y^5} }, (C) 2y5{ 2y^5 }, (D) 4y5{ 4y^5 }.

Identifying the Corrected Error and Re-Simplifying

Let's revisit each step meticulously to find where the error occurred. We'll start from the beginning:

  1. Apply the Power of a Quotient:

    (16y8)14=1614(y8)14{ \left(\frac{16}{y^8}\right)^{\frac{1}{4}} = \frac{16^{\frac{1}{4}}}{(y^8)^{\frac{1}{4}}} }

    This step is correct.

  2. Simplify the Numerator:

    1614=164=2{ 16^{\frac{1}{4}} = \sqrt[4]{16} = 2 }

    This step is also correct.

  3. Simplify the Denominator:

    (y8)14=y8Γ—14=y2{ (y^8)^{\frac{1}{4}} = y^{8 \times \frac{1}{4}} = y^2 }

    This step is correct as well.

  4. Substitute Simplified Terms Back into the Expression:

    1614(y8)14=2y2{ \frac{16^{\frac{1}{4}}}{(y^8)^{\frac{1}{4}}} = \frac{2}{y^2} }

    Correct substitution.

  5. Multiply by y3{ y^3 }:

    Here's where we need to be extra careful:

    2y2Γ—y3=2Γ—y3y2{ \frac{2}{y^2} \times y^3 = 2 \times \frac{y^3}{y^2} }

    This setup is correct.

  6. Final Simplification (Corrected):

    Now, let's simplify the exponents correctly:

    y3y2=y3βˆ’2=y1=y{ \frac{y^3}{y^2} = y^{3-2} = y^1 = y }

    So, the expression simplifies to:

    2Γ—y=2y{ 2 \times y = 2y }

Still, this result (2y{ 2y }) doesn't match any of the given options. We need to thoroughly re-examine the calculations to find the discrepancy.

Unveiling the Persistent Error

Okay, let's take a step back and look at the entire process again. Sometimes, a fresh perspective can help spot a mistake that's been overlooked. We have the expression:

(16y8)14Γ—y3{ \left(\frac{16}{y^8}\right)^{\frac{1}{4}} \times y^3 }

Let’s retrace our steps with extra caution:

  1. Power of a Quotient:

    (16y8)14=1614(y8)14{ \left(\frac{16}{y^8}\right)^{\frac{1}{4}} = \frac{16^{\frac{1}{4}}}{(y^8)^{\frac{1}{4}}} }

    This is correct.

  2. Simplify Numerator:

    1614=164=2{ 16^{\frac{1}{4}} = \sqrt[4]{16} = 2 }

    Correct.

  3. Simplify Denominator:

    (y8)14=y8Γ—14=y2{ (y^8)^{\frac{1}{4}} = y^{8 \times \frac{1}{4}} = y^2 }

    Also correct.

  4. Substitute Back:

    2y2Γ—y3{ \frac{2}{y^2} \times y^3 }

    Correct substitution.

  5. Multiply and Simplify:

    2y2Γ—y3=2y3βˆ’2=2y{ \frac{2}{y^2} \times y^3 = 2y^{3-2} = 2y }

    The error lies here! We keep getting 2y{ 2y }, which isn't an option. This suggests there might be a misunderstanding of the final form required or a mistake in the provided options themselves.

Final Conclusion Based on Steps:

Given our meticulous step-by-step simplification, the correct answer should indeed be 2y{ 2y }. However, since this is not among the provided options, it's crucial to consider that there might be an error in the question itself or in the multiple-choice answers provided. If forced to choose from the given options, it would be incorrect to select any of them, as they do not match the correct simplification.

Analysis of the Answer Choices

Let’s analyze the given answer choices:

A) 2y5{ \frac{2}{y^5} }

B) 4y5{ \frac{4}{y^5} }

C) 2y5{ 2y^5 }

D) 4y5{ 4y^5 }

None of these options match our simplified expression, which is 2y{ 2y }. This further supports the conclusion that there may be an error in the question or the answer choices. In a real test scenario, it would be important to double-check the original problem and, if possible, consult with the instructor about the discrepancy.

Final Thoughts

Simplifying expressions requires careful application of exponent rules and algebraic principles. In this detailed walkthrough, we methodically simplified the given expression, ensuring each step was justified by fundamental mathematical properties. While our calculated result, 2y{ 2y }, does not align with the provided answer choices, the process underscores the importance of accuracy and thoroughness in algebraic manipulations. If faced with such a situation in an exam, it would be prudent to verify the question and the options or seek clarification from the instructor.

Understanding the process is as crucial as arriving at the correct answer. This exercise reinforces the significance of step-by-step problem-solving, meticulous application of mathematical rules, and critical analysis of results. By mastering these skills, you can confidently approach a wide range of algebraic challenges.