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Chapter 3: Problem 3

Evaluate the expression. a. \(9(9)^{-1 / 2}\) b. \(5(5)^{-1 / 2}\)

Short Answer

Expert verified
The given expressions can be evaluated as follows: a. \(9(9)^{-1/2} = 3\) b. \(5(5)^{-1/2} \approx 2.23\)

Step by step solution

01

Understand the rational exponent

The exponent -1/2 can be written as a combination of two exponents: -1 and 1/2. When an exponent is -1, it represents taking the reciprocal (i.e., flipping) of the number. When the exponent is 1/2, it represents taking the square root of the number. So, to evaluate the expression \(9(9)^{-1/2}\), we will first find the square root and then find the reciprocal of the result.
02

Apply the exponent rules

We start by taking the square root of 9 and then finding the reciprocal of the result. Square root of 9: \(\sqrt{9}=3\) Reciprocal of the result: \(\frac{1}{3}\) Now we multiply 9 by the reciprocal of the result: \(9(\frac{1}{3}) = 3\) The expression a. \(9(9)^{-1/2}\) simplifies to 3. For expression b. \(5(5)^{-1/2}\):
03

Understand the rational exponent

As explained before, the exponent -1/2 can be written as a combination of two exponents: -1 and 1/2. When an exponent is -1, it represents taking the reciprocal (i.e., flipping) of the number. When the exponent is 1/2, it represents taking the square root of the number. So, to evaluate the expression \(5(5)^{-1/2}\), we will first find the square root and then find the reciprocal of the result.
04

Apply the exponent rules

We start by taking the square root of 5 and then finding the reciprocal of the result. Square root of 5 (rounded to two decimal places): \(\sqrt{5} \approx 2.24\) Reciprocal of the result: \(\frac{1}{2.24} \approx 0.446\) Now we multiply 5 by the reciprocal of the result: \(5(0.446) \approx 2.23\) The expression b. \(5(5)^{-1/2}\) simplifies to approximately 2.23.

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Key Concepts

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

Reciprocal
The reciprocal of a number is basically its flipped version. To find the reciprocal of a number, you swap the numerator with the denominator. If the number is a whole number, you can think of it as being over 1 before flipping it upside down. For example, the reciprocal of 3 is \( \frac{1}{3} \). This concept is handy when dealing with negative exponents.
  • When you see \( a^{-1} \), it is the same as \( \frac{1}{a} \).
  • In the exercise, the exponent \(-1/2\) means you'll first find the reciprocal after taking the square root.
Understanding reciprocals can help simplify calculations involved in algebra, fractions, and when dealing with rational exponents.
Square Root
A square root of a number is essentially asking "what number multiplied by itself gives me this number?". Square roots are often represented by the radical symbol \( \sqrt{} \). For example, the square root of 9 is 3, since \(3 \times 3 = 9\).
  • In the expressions \(9^{-1/2}\) and \(5^{-1/2}\), the \( \frac{1}{2} \) exponent signifies that we need to find the square root first.
  • This is crucial because it transforms the original number into one that can be easily flipped as a reciprocal.
Once the square root is found, further calculations become simpler, especially when combined with exponent rules.
Exponent Rules
Exponent rules are guidelines that make handling exponents manageable, particularly with rational exponents. Some key rules include:
  • Product of Powers: \( a^m \times a^n = a^{m+n} \)
  • Power of a Power: \( (a^m)^n = a^{m \times n} \)
  • Negative Exponent: \( a^{-m} = \frac{1}{a^m} \)
  • Rational Exponent: \( a^{\frac{m}{n}} = \sqrt[n]{a^m} \)
For rational exponents, the fraction describes two operations: the numerator is the power, and the denominator is the root. For instance, in the exercises, \(-1/2\) suggests performing a square root operation followed by taking the reciprocal of the result.Understanding and applying these rules can simplify complex expressions, making them more approachable and easier to solve.

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Most popular questions from this chapter

Sketch the graphs of the given functions on the same axes. \(y=1-e^{-x}\) and \(y=1-e^{-0.5 x}\)

The number of Internet users in China is projected to be $$ N(t)=94.5 e^{0.2 t} \quad(1 \leq t \leq 6) $$ where \(N(t)\) is measured in millions and \(t\) is measured in years, with \(t=1\) corresponding to the beginning of 2005 . a. How many Internet users were there at the beginning of \(2005 ?\) At the beginning of 2006 ? b. How many Internet users are there expected to be at the beginning of 2010 ? c. Sketch the graph of \(N\).

The percentage of families that were married households between 1970 and 2000 is approximately $$ P(t)=86.9 e^{-0.05 t} \quad(0 \leq t \leq 3) $$ where \(t\) is measured in decades, with \(t=0\) corresponding to the beginning of 1970 . a. What percentage of families were married households at the beginning of \(1970,1980,1990\), and 2000 ? b. Sketch the graph of \(P\).

Use the laws of logarithms to expand and simplify the expression. $$\ln \frac{e^{x}}{1+e^{x}}$$

Given that \(\log 3 \approx 0.4771\) and \(\log 4 \approx\) 0.6021, find the value of each logarithm. $$\log \frac{1}{300}$$
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