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Chapter 2: Problem 123

Simplify the following problems. $$ \left(x^{5 t} y^{4 r}\right)^{7} $$

Short Answer

Expert verified
Answer: The simplified form of the expression is \(x^{35t} \cdot y^{28r}\).

Step by step solution

01

Identify the exponents

We are given the expression \((x^{5t} y^{4r})^7\). To simplify this expression, we will need to apply the power-rule for exponents, which states \((a^m \cdot b^n)^p = a^{mp} \cdot b^{np}\). In this case, \(a=x^{5t}\), \(b=y^{4r}\), and \(p=7\).
02

Apply the power rule

Applying the power-rule, we can simplify the expression to: $$ (x^{5t} y^{4r})^7 = x^{(5t)(7)} \cdot y^{(4r)(7)} $$
03

Perform the multiplication

Now, we multiply the exponents by 7: $$ x^{(5t)(7)} \cdot y^{(4r)(7)} = x^{35t} \cdot y^{28r} $$
04

Write the simplified expression

Finally, we have the simplified expression: $$ (x^{5t} y^{4r})^7 = x^{35t} \cdot y^{28r} $$

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Key Concepts

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

Understanding the Power-Rule
The power-rule is a fundamental principle in managing exponents. It assists in simplifying expressions involving exponents raised to another power. When you see an expression like \((a^m \cdot b^n)^p\), the rule tells you to distribute the outer power to each part inside. So, it becomes \(a^{m \cdot p} \cdot b^{n \cdot p}\).

This rule is very handy when dealing with powers inside parentheses. Here鈥檚 a breakdown:
  • Distribute the exponent: Apply the outer exponent to every base inside the parenthesis.
  • Multiply exponents: Each base retains their internal structure, but the exponent is multiplied by that outer number.
By using the power-rule, you can simplify expressions and make calculations faster and error-free.

In our given problem, applying the power-rule to \((x^{5t} y^{4r})^7\) breaks it into smaller, more manageable pieces.
Importance of Simplification
Simplification is key in algebra to make expressions easier to read and work with. It means shrinking down complex expressions into simpler ones.

Why simplify?
  • Easier understanding: Simpler forms are easier to understand and solve.
  • Reduces error: Less clutter reduces the chance of mistakes in later steps.
The goal is to get your expression into a basic form, following mathematical rules closely.

When handling expressions like \((x^{5t} y^{4r})^7\), simplification helps us see the underlying structure. You end with \(x^{35t} \cdot y^{28r}\) which is much tidier. Simplicity is the heart of mathematical elegance!
Mastering Multiplication of Exponents
Multiplying exponents is about recognizing patterns and applying rules systematically. Whenever you handle terms like \(a^m \cdot a^n\), you add the exponents: \(a^{m + n}\). However, what about exponents raised further by another exponent? That's where our exercise and learned rules play in.

For expressions such as \(x^{5t} \cdot y^{4r}\) raised to a power, each exponent is multiplied by the exterior exponent. For instance:
  • \(x^{5t} \rightarrow x^{5t \cdot 7} = x^{35t}\)
  • \(y^{4r} \rightarrow y^{4r \cdot 7} = y^{28r}\)

This multiplication of exponents simplifies the calculation and reinforces understanding of how quantities interact in exponential terms. Understanding these rules makes handling powers in mathematics much less daunting.

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Most popular questions from this chapter

For the following problems, use the distributive property to expand the quantities.

Simplify the following problems. $$ x^{4} \cdot \frac{x^{10}}{x^{3}} $$

For the following problems, write the appropriate relation symbol \((=,<,>)\). $$ -8 \quad-5 $$

Find the value of \(\frac{4^{2}+(3+2)^{2}-1}{2^{3} \cdot 5}+\frac{2^{4}\left(3^{2}-2^{3}\right)}{4^{2}}\).

For the following problems, what integers can replace \(x\) so that the statements are true? $$ 4 \leq x \leq 7 $$
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