/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Write each expression in power f... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write each expression in power form \(a x^{b}\) for numbers \(a\) and \(b\). $$ \frac{24}{(2 \sqrt{x})^{3}} $$

Short Answer

Expert verified
The expression in power form is \(3x^{-3/2}\).

Step by step solution

01

Simplify the Denominator

First, we simplify the denominator \((2 \sqrt{x})^3\). Recognize that \(\sqrt{x} = x^{1/2}\), so \(2\sqrt{x}\) can be rewritten as \(2x^{1/2}\). Therefore, \((2\sqrt{x})^3 = (2x^{1/2})^3\).
02

Expand the Exponent

We know that \((ab)^n = a^n \cdot b^n\). Apply this to \((2x^{1/2})^3 = 2^3 \cdot (x^{1/2})^3\). Calculate these separately: \(2^3 = 8\) and \((x^{1/2})^3 = x^{3/2}\). Combine to get: \(8x^{3/2}\).
03

Express the Original Expression

Now write the original expression \(\frac{24}{(2 \sqrt{x})^{3}}\) as \(\frac{24}{8x^{3/2}}\). Simplify this by dividing both the numerator and the denominator by 8, resulting in \(\frac{3}{x^{3/2}}\).
04

Convert to Power Form

The expression \(\frac{3}{x^{3/2}}\) can be rewritten as \(3x^{-3/2}\) by recognizing \(\frac{1}{x^{n}} = x^{-n}\). Thus, the expression in power form is \(3x^{-3/2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Exponents
Exponents are a shorthand way of expressing repeated multiplication of a number by itself. The expression of a number, say \(a\), raised to a power \(n\) is written as \(a^n\). This indicates that \(a\) is multiplied by itself \(n\) times. Understanding exponents is crucial in algebra because they simplify complex expressions and calculations.

Let's look at some examples:
  • \(2^3 = 2 \times 2 \times 2 = 8\)
  • \(5^2 = 5 \times 5 = 25\)
  • \(x^4 = x \times x \times x \times x\)
Exponents follow specific rules that make them useful, such as:
  • Product of Powers: \(a^m \cdot a^n = a^{m+n}\)
  • Power of a Power: \((a^m)^n = a^{m \cdot n}\)
  • Power of a Product: \((ab)^n = a^n \cdot b^n\)
These rules help simplify expressions, especially when dealing with complex algebraic problems like simplifying \((2x^{1/2})^3\).
Simplifying Expressions
Simplifying expressions means transforming them into a more basic or digestible form while still having the same value. One of the methods frequently used in simplification is applying the rules of exponents.

For the example in our original exercise, simplification involved several steps:
First, we needed to simplify the denominator \((2\sqrt{x})^3\), rewritten as \((2x^{1/2})^3\) after identifying \(\sqrt{x}\) as \(x^{1/2}\). Next, applying the Power of a Product rule, we expanded this expression into \(2^3 \cdot (x^{1/2})^3\), which computes to \(8x^{3/2}\).

An essential aspect of simplification is realizing that the rules provide a systematic method to break down expressions into elementary parts. Dividing and reducing fractions, as we did with \(\frac{24}{8x^{3/2}}\) to \(\frac{3}{x^{3/2}}\), is also part of simplification, bringing expressions into their lowest terms for easier computation and understanding.
Fractional Exponents
Fractional exponents represent roots, and understanding them is key to manipulating algebraic expressions in power form. The notation \(x^{m/n}\) signifies the \(n\)-th root of \(x\) raised to the \(m\)-th power. For example, \(x^{1/2}\) means the square root of \(x\), while \(x^{3/2}\) indicates that you first take the square root of \(x\) and then cube the result.

In algebra, fractional exponents provide a more unified and powerful method to express roots alongside integer exponents. They allow you to simplify expressions like radicals, making calculations more manageable and systematic.

Returning to our exercise, we converted \(\frac{3}{x^{3/2}}\) to the power form \(3x^{-3/2}\) by applying the rule that \(\frac{1}{x^n} = x^{-n}\). This conversion to negative exponents shows how flexible and valuable fractional exponents are, enabling easier handling of division or root operations in power form.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

GENERAL: Seat Belt Use Because of driver education programs and stricter laws, seat belt use has increased steadily over recent decades. The following table gives the percentage of automobile occupants using seat belts in selected years. $$ \begin{array}{lcccc} \hline \text { Year } & 1995 & 2000 & 2005 & 2010 \\ \hline \text { Seat Belt Use (\%) } & 60 & 71 & 81 & 86 \\ \hline \end{array} $$ a. Number the data columns with \(x\) -values \(1-4\) and use linear regression to fit a line to the data. State the regression formula. [Hint: See Example 8.] b. Interpret the slope of the line. From your answer, what is the yearly increase? c. Use the regression line to predict seat belt use in \(2015 .\) d. Would it make sense to use the regression line to predict seat belt use in 2025 ? What percentage would you get?

The following function expresses an income tax that is \(10 \%\) for incomes below \(\$ 5000\), and otherwise is \(\$ 500\) plus \(30 \%\) of income in excess of \(\$ 5000\). \(f(x)=\left\\{\begin{array}{ll}0.10 x & \text { if } 0 \leq x<5000 \\\ 500+0.30(x-5000) & \text { if } x \geq 5000\end{array}\right.\) a. Calculate the tax on an income of \(\$ 3000\). b. Calculate the tax on an income of \(\$ 5000\). c. Calculate the tax on an income of \(\$ 10,000\). d. Graph the function.

The intersection of an isocost line \(w L+r K=C\) and an isoquant curve \(K=a L^{b}\) (see pages 18 and 32 ) gives the amounts of labor \(L\) and capital \(K\) for fixed production and cost. Find the intersection point \((L, K)\) of each isocost and isoquant. [Hint: After substituting the second expression into the first, multiply through by \(L\) and factor.] $$ 5 L+4 K=120 \text { and } K=180 \cdot L^{-1} $$

63-64. BUSINESS: Straight-Line Depreciation Straight-line depreciation is a method for estimating the value of an asset (such as a piece of machinery) as it loses value ( \({ }^{\prime \prime}\) depreciates" \()\) through use. Given the original price of an asset, its useful lifetime, and its scrap value (its value at the end of its useful lifetime), the value of the asset after \(t\) years is given by the formula: $$ \begin{aligned} \text { Value }=(\text { Price })-&\left(\frac{(\text { Price })-(\text { Scrap value })}{(\text { Useful lifetime })}\right) \cdot t \\ & \text { for } 0 \leq t \leq(\text { Useful lifetime }) \end{aligned} $$ a. A farmer buys a harvester for $$\$ 50,000$$ and estimates its useful life to be 20 years, after which its scrap value will be $$\$ 6000$$. Use the formula above to find a formula for the value \(V\) of the harvester after \(t\) years, for \(0 \leq t \leq 20\). b. Use your formula to find the value of the harvester after 5 years. c. Graph the function found in part (a) on a graphing calculator on the window \([0,20]\) by \([0,50,000]\). [Hint: Use \(x\) instead of \(t\).]

Use the TABLE feature of your graphing calculator to evaluate \(\left(1+\frac{1}{x}\right)^{x}\) for values of \(x\) such as \(100,10,000,1,000,000\), and higher values. Do the resulting numbers seem to be approaching a limiting value? Estimate the limiting value to five decimal places. The number that you have approximated is denoted \(e\), and will be used extensively in Chapter \(4 .\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.