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Chapter 1: Problem 77

Simplify. $$ \left(3 x^{2} y^{5} z\right)^{3} $$

Short Answer

Expert verified
The simplified expression is \(27x^6y^{15}z^3\).

Step by step solution

01

Apply the Power to Each Factor Individually

When a product inside the parentheses is raised to a power, apply that power to each factor inside the parentheses. The expression is \((3 x^2 y^5 z)^3\). So, each element inside will be raised to the 3rd power: \(3^3\), \((x^2)^3\), \((y^5)^3\), and \(z^3\).
02

Calculate the Coefficient Raised to the Power

Calculate \(3^3\). This is simply \(3 \times 3 \times 3 = 27\).
03

Apply the Power to the Variables

Raise each variable to the power of 3. For \((x^2)^3\), use the power of a power rule, \((a^m)^n = a^{m \cdot n}\). Thus, \((x^2)^3 = x^{2 \cdot 3} = x^6\). Similarly, \((y^5)^3 = y^{5 \cdot 3} = y^{15}\) and \(z^3 = z^3\).
04

Combine All Parts

Combine all elements calculated: \(27 x^6 y^{15} z^3\). This gives the final simplified form of the expression.

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Key Concepts

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

Exponentiation Rules
Exponentiation rules are essential when working with powers and exponents in algebra. These rules help simplify expressions that involve exponential terms. There are a few principles you should always remember.

  • Product of Powers Rule: When multiplying with the same base, add the exponents. For example, if you have \(a^m \times a^n\), this simplifies to \( a^{m+n} \).
  • Power of a Power Rule: When raising an exponent to another power, multiply the exponents: \((a^m)^n = a^{m \cdot n}\).
  • Power of a Product Rule: When raising a product to a power, apply the exponent to each factor separately: \((ab)^n = a^n \times b^n\).

These basic rules apply to any real numbers and allow for effective simplification of even complex algebraic expressions. Grasping these concepts will make solving exercises involving powers much more straightforward.
Power of a Product Rule
The Power of a Product Rule is a technique used to handle expressions where multiple different factors are raised to a power. This rule can often simplify products that involve exponents. Suppose you have an expression such as \((ab)^n\).

This is handled by applying the exponent \(n\) to each factor inside the bracket individually, resulting in \(a^n \times b^n\). This method ensures each component of the product is elevated to the specified power.

For example, consider the expression \((3x^2y^5z)^3\):
  • Apply the power 3 to the coefficient: \(3^3 = 27\).
  • Apply the power to \(x^2\), giving \((x^2)^3 = x^{6}\).
  • Apply the power to \(y^5\), giving \((y^5)^3 = y^{15}\).
  • Finally, address \(z\), giving \(z^3\).

By using the Power of a Product Rule, you maintain clarity and simplicity while analyzing and simplifying expressions with multiple factors.
Simplifying Expressions
Simplifying expressions involving exponents is a key skill in algebra. It involves reducing expressions to their simplest form, making them easier to work with. Here are the foundational steps:

  • Identify and group similar terms and variables within the expression.
  • Apply relevant exponentiation rules, such as the Power of a Product Rule, to manage powers efficiently.
  • Calculate constants separately, ensuring clarity and correctness.
  • Combine the terms into a final, simplified expression.

Consider the expression \((3x^2y^5z)^3\). Each variable and number is addressed using the applicable rules:
  • First, calculate \(3^3\), resulting in 27.
  • Using the Power Rule, combine the exponents of the variable \(x\) to get \(x^6\).
  • Do the same for \(y\), resulting in \(y^{15}\).
  • Apply the power to \(z\), yielding \(z^3\).

The final expression, then, is neatly organized as \(27x^6 y^{15} z^3\). Learning and using these steps empowers you to simplify expressions effortlessly, ensuring accuracy in more complex algebraic problems.

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Most popular questions from this chapter

\(87-88 .\) ALLOMETRY: Dinosaurs The study of size and shape is called "allometry," and many allometric relationships involve exponents that are fractions or decimals. For example, the body measurements of most four-legged animals, from mice to elephants, obey (approximately) the following power law: $$ \left(\begin{array}{c} \text { Average body } \\ \text { thickness } \end{array}\right)=0.4 \text { (hip-to-shoulder length) }^{3 / 2} $$ where body thickness is measured vertically and all measurements are in feet. Assuming that this same relationship held for dinosaurs, find the average body thickness of the following dinosaurs, whose hip-toshoulder length can be measured from their skeletons: Diplodocus, whose hip-to-shoulder length was 16 feet.

Simplify. $$ \left[\left(x^{2}\right)^{2}\right]^{2} $$

GENERAL: Earthquakes The sizes of major earthquakes are measured on the Moment Magnitude Scale, or MMS, although the media often still refer to the outdated Richter scale. The MMS measures the total energy relensed by an earthquake, in units denoted \(M_{W}\) (W for the work accomplished). An increase of \(1 M_{W}\) means the energy increased by a factor of \(32,\) so an increase from \(A\) to \(B\) means the energy increased by a factor of \(32^{B-A}\). Use this formula to find the increase in energy between the following earthquakes: The 2001 earthquake in India that measured \(7.7 M_{W}\) and the 2011 earthquake in Japan that measured \(9.0 M_{W}\). (The earthquake in Japan generated a 28 -foot tsunami wave that traveled six miles inland, killing 24,000 and causing an estimated \(\$ 300\) billion in damage, making it the most expensive natural disaster ever recorded.)

If a linear function is such that \(f(2)=5\) and \(f(3)=7,\) then \(f(4)=?\) [Hint: No work necessary. \(]\)

Simplify. $$ \frac{\left(u^{3} v w^{2}\right)^{2}}{9\left(u^{2} w\right)^{2}} $$
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