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Chapter 1: Problem 25

Evaluate each expression without using a calculator. $$ (-8)^{5 / 3} $$

Short Answer

Expert verified
(-8)^{5/3} = -32.

Step by step solution

01

Simplify the Fractional Exponent

The expression \[(-8)^{5/3}\]can be broken down using the fractional exponent rule. This means we can express it as \[((-8)^{1/3})^5\].The base \(-8\) is being raised to the \(1/3\) power first, and then the result is raised to the \(5^{th}\) power.
02

Evaluate the Cube Root

Now calculate \((-8)^{1/3}\), which is the cube root of \(-8\). The cube root of \(-8\) is \(-2\) because \((-2) \times (-2) \times (-2) = -8\). Therefore, \((-8)^{1/3} = -2\).
03

Raise the Result to the Fifth Power

Now raise the result from Step 2 to the power of 5:\[(-2)^5\].Calculate \((-2) \times (-2) \times (-2) \times (-2) \times (-2)\). That's \(32\) because\[-2 \times -2 = 4, \]\[4 \times -2 = -8, \]\[-8 \times -2 = 16, \]\[16 \times -2 = -32.\]Therefore, \((-2)^5 = -32\).

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Key Concepts

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

Simplifying Expressions
Simplifying expressions can often make many mathematical problems easier to understand and solve. In this context, we are dealing with fractional exponents. Fractional exponents indicate both a root and a power. For example, in the expression \((-8)^{5/3}\), the exponent \(5/3\) signifies two operations:
  • The denominator: 3 signifies that we first need to find the cube root of the base number \(-8\).
  • The numerator: 5 indicates that we then raise the result of the cube root to the power of 5.
Breaking down fractional exponents into roots and powers simplifies complex expressions, allowing us to solve them in manageable steps. This not only aids in computation but also allows a clearer understanding of the operations involved.
Cube Root Calculation
Cube root calculation is a fundamental skill in understanding fractional exponents. The cube root of a number \(x\) is what you would multiply by itself twice to return to the original number \(x\). More formally, the cube root of a number is written as \(x^{1/3}\).
In our example, we look at the expression \((-8)^{1/3}\). This tells us to find what number, when cubed, gives us \(-8\). In this case, the answer is \(-2\). Why? Because:
  • \((-2) \times (-2) \times (-2) = -8\).
Understanding cube roots is crucial for dealing with expressions where numbers are raised to fractional powers. This skill helps unravel complex expressions into more straightforward, manageable tasks.
Power of a Negative Number
Working with powers of a negative number is another important area in mathematics. Negative numbers operate under unique rules when involving exponents. The key is to pay attention to whether the exponent is even or odd:
  • Odd powers: The result remains negative. For example, \((-2)^5\) remains \(-32\).
  • Even powers: The result becomes positive. For example, \((-2)^4 = 16\).
In our calculation, once the cube root of \(-8\) was determined to be \(-2\), we then raised it to the fifth power: \((-2)^5\). With five as an odd number, the final result remains negative. Hence, calculating powers of negative numbers requires attention to such details, making sure that multiplication rules are correctly applied.

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Most popular questions from this chapter

Wind power, popular for environmental and energy security reasons, is growing rapidly in the United States and Europe. Total world wind energy-generating capacity, in megawatts, is given in the following table. (One megawatt would power about 1000 average homes.) $$ \begin{array}{l|l|l|l|l} \hline \text { Year } & 1990 & 1995 & 2000 & 2005 \\ \hline \begin{array}{l} \text { Wind-Generating } \\ \text { Capacity (megawatts) } \end{array} & 1930 & 4780 & 18,450 & 59,091 \\ \hline \end{array} $$ a. Number the data columns with \(x\) -values \(0-3\) (so that \(x\) stands for the number of five-year intervals since 1990 ) and use exponential regression to fit a curve to the data. State the regression formula. [Hint: See Example 11.] b. Use the regression function to predict world wind-energy capacity in the year \(2015 .\)

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 .

Find, rounding to five decimal places: a. \(\left(1+\frac{1}{100}\right)^{100}\) b. \(\left(1+\frac{1}{10,000}\right)^{10,000}\) c. \(\left(1+\frac{1}{1,000,000}\right)^{1,000,000}\) d. 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 .

Graph the parabola \(y_{1}=1-x^{2}\) and the semicircle \(y_{2}=\sqrt{1-x^{2}}\) on the window \([-1,1]\) by \([0,1] .\) (You may want to adjust the window to make the semicircle look more like a semicircle.) Use TRACE to determine which is the "inside" curve (the parabola or the semicircle) and which is the "outside" curve. These graphs show that when you graph a parabola, you should draw the curve near the vertex to be slightly more "pointed" than a circular curve.

The following problems extend and augment the material presented in the text. For any \(x\), the function \(\operatorname{INT}(x)\) is defined as the greatest integer less than or equal to \(x\). For example, \(\operatorname{INT}(3.7)=3\) and \(\operatorname{INT}(-4.2)=-5\) a. Use a graphing calculator to graph the function \(y_{1}=\operatorname{INT}(x)\). (You may need to graph it in DOT mode to eliminate false connecting lines.) b. From your graph, what are the domain and range of this function?
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