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Chapter 8: Problem 147

Simplify. (a) \(32^{\frac{2}{5}}\) (b) \(27^{-\frac{2}{3}}\)(c) \((-25)^{\frac{1}{2}}\)

Short Answer

Expert verified
(a) 4, (b) \(\frac{1}{9}\), (c) 5i

Step by step solution

01

Simplify Part (a)

Given expression: \[ 32^{\frac{2}{5}} \]First, rewrite 32 as a base of 2: \[ 32 = 2^5 \]So the expression becomes: \[ (2^5)^{\frac{2}{5}} \]Apply the power rule of exponents \[ (a^m)^n = a^{m*n} \]Thus:\[ (2^5)^{\frac{2}{5}} = 2^{5 \cdot \frac{2}{5}} = 2^2 = 4 \]So, \[ 32^{\frac{2}{5}} = 4 \]
02

Simplify Part (b)

Given expression: \[ 27^{-\frac{2}{3}} \]First, rewrite 27 as a base of 3: \[ 27 = 3^3 \]So the expression becomes: \[ (3^3)^{-\frac{2}{3}} \]Apply the power rule of exponents\[ (a^m)^n = a^{m*n} \]Thus:\[ (3^3)^{-\frac{2}{3}} = 3^{3 \cdot (-\frac{2}{3})} = 3^{-2} \]Use the negative exponent rule:\[ a^{-m} = \frac{1}{a^m} \]Thus:\[ 3^{-2} = \frac{1}{3^2} = \frac{1}{9} \]So, \[ 27^{-\frac{2}{3}} = \frac{1}{9} \]
03

Simplify Part (c)

Given expression: \[ (-25)^{\frac{1}{2}} \]The exponent \(\frac{1}{2}\) indicates taking the square root. Therefore, \[ (-25)^{\frac{1}{2}} = \sqrt{-25} \]Since the square root of a negative number is not a real number (it falls into the category of imaginary numbers), \[ \sqrt{-25} = 5i \]where \(i\) is the imaginary unit defined as \(i^2 = -1\).

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Key Concepts

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

Power Rule of Exponents
Understanding the power rule of exponents is essential when simplifying expressions. The rule states that \((a^m)^n = a^{m*n}\). This means when raising a power to another power, you multiply the exponents.
Let's look at an example: \((2^5)^{2/5}\). First, we rewrite 32 as \(2^5\), so the expression becomes \((2^5)^{2/5}\). Using the power rule, \((2^5)^{2/5} = 2^{5*(2/5)} = 2^2 = 4\). This simplifies \(32^{2/5}\) to 4.
Understanding this rule can significantly help in simplifying expressions with exponents.
Negative Exponents
Negative exponents may seem tricky, but they follow a simple rule: \(a^{-m} = \frac{1}{a^m}\). This means a negative exponent indicates taking the reciprocal of the base raised to the positive exponent.
Consider the expression \(27^{-2/3}\):
  • First, rewrite 27 as \(3^3\), making the expression \((3^3)^{-2/3}\).
  • Using the power rule again: \((3^3)^{-2/3} = 3^{3*(-2/3)} = 3^{-2}\).
  • Apply the negative exponent rule: \(3^{-2} = \frac{1}{3^2} = \frac{1}{9}\).
Thus, \(27^{-2/3} = \frac{1}{9}\).
Imaginary Numbers
Imaginary numbers can come into play when dealing with square roots of negative numbers. The basic imaginary unit is \(i\), which is defined such that \(i^2 = -1\).
Consider the expression \((-25)^{1/2}\):
  • The exponent \(1/2\) indicates taking the square root, so we have \(\sqrt{-25}\).
  • This involves the square root of a negative number. Since \(\sqrt{-25}\) is not a real number, it is an imaginary number.
  • Thus, \(\sqrt{-25} = 5i\), where \(i\) is the imaginary unit.
So, \((-25)^{1/2} = 5i\). Imaginary numbers like \(i\) enable us to work with solutions that are not real numbers.

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Most popular questions from this chapter

Simplify. (a) \((-32)^{\frac{1}{5}}\) (b) \((243)^{-\frac{1}{5}}\) (c)\(-125^{\frac{1}{3}}\)

Simplify. (a) \(81^{\frac{1}{2}}\) (b) \(125^{\frac{1}{3}}\) (c) \(64^{\frac{1}{2}}\)

Use the Quotient Property to simplify square roots. (a) \(\sqrt{\frac{50 r^{5} s^{2}}{128 r^{2} s^{6}}}\) \\}(b) \(\sqrt[3]{\frac{24 m^{9} n^{7}}{375 m^{4} n}}\)(c) \(\sqrt[4]{\frac{81 m^{2} n^{8}}{256 m^{1} n^{2}}}\)

Use the Quotient Property to simplify square roots. (a) \(\sqrt{\frac{27 p^{2} q}{108 p^{4} q^{3}}}\) (b) \(\sqrt[3]{\frac{16 c^{5} d^{7}}{250 c^{2} d^{2}}}\)(c) \(\sqrt[6]{\frac{2 m^{9} n^{7}}{128 m^{3} n}}\)

Simplify using absolute values as necessary. (a) \(\sqrt[7]{128 r^{14}}\) (b) \(\sqrt[4]{81 s^{24}}\)
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