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Chapter 10: Problem 38

Rewrite with a positive exponent and evaluate. \(\left(\frac{1}{32}\right)^{-1 / 5}\)

Short Answer

Expert verified
\(\left(\frac{1}{32}\right)^{-1 / 5} = 2\)

Step by step solution

01

Apply Negative Exponent Rule

Recall that the negative exponent rule states that \(a^{-n} = \frac{1}{a^{n}}\). In this case, we have \(\left(\frac{1}{32}\right)^{-1/5}\). Applying the negative exponent rule, we have the following: \[ \left(\frac{1}{32}\right)^{-1/5} = \frac{1}{\left(\left(\frac{1}{32}\right)^{1/5}\right)} \]
02

Apply Rational Exponent Rule

Now, apply the rational exponent rule, which states that \(a^{m/n} = \sqrt[n]{a^m}\). In this case, we have the following: \[ \frac{1}{\left(\left(\frac{1}{32}\right)^{1/5}\right)} = \frac{1}{\sqrt[5]{\left(\frac{1}{32}\right)}} \]
03

Rewriting expression

Next, we will rewrite "\(\frac{1}{32}\)" as \(2^{-5}\) so that it's easier to evaluate: \[ \frac{1}{\sqrt[5]{\left(\frac{1}{32}\right)}} = \frac{1}{\sqrt[5]{(2^{-5})}} \]
04

Evaluate 5th Root

Now, using the properties of exponents within a root, i.e., \(\sqrt[n]{a^m} = a^{\frac{m}{n}}\), we have: \[ \frac{1}{\sqrt[5]{(2^{-5})}} =\frac{1}{2^{-1}} \]
05

Apply Negative Exponent Rule Again

Once again, apply the negative exponent rule, which states that \(a^{-n} = \frac{1}{a^{n}}\). In this case, we have the following: \[ \frac{1}{2^{-1}} = 2^1 \]
06

Evaluate Result

Finally, we have reached an expression that can be easily evaluated: \[ 2^1 = 2 \] So, \(\left(\frac{1}{32}\right)^{-1 / 5} = 2\).

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Key Concepts

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

Understanding Rational Exponents
When you see an expression like \(a^{m/n}\), you're looking at a rational exponent. Rational exponents are a way to represent roots and powers in one expression. Here鈥檚 what each part means:
  • \(m\) is the power (or the numerator of the exponent).
  • \(n\) is the root (or the denominator of the exponent).
So \(a^{m/n}\) is equivalent to the \(n\)-th root of \(a\) raised to the \(m\)-th power: \(\sqrt[n]{a^m}\). This concept allows us to handle both powers and roots efficiently. For example, when dealing with \(\left(\frac{1}{32}\right)^{-1/5}\), the rational exponent \(-1/5\) indicates taking the 5th root and then dealing with the inverse.
Mastering Exponent Rules
Exponent rules simplify the way we handle expressions involving powers. Here are some key rules to remember:
  • Negative Exponent Rule: \(a^{-n} = \frac{1}{a^n}\). This helps convert negative exponents into positive ones by creating a reciprocal.
  • Product of Powers: \(a^m \times a^n = a^{m+n}\). This combines similar bases.
  • Power of a Power: \((a^m)^n = a^{m \times n}\). This multiplies the exponents.
In the given problem, we used the Negative Exponent Rule to rewrite \(\left(\frac{1}{32}\right)^{-1/5}\). It transformed into a more manageable positive exponent, facilitating further simplification.
Evaluating Roots
Root evaluation is a critical step when working with rational exponents. Here, we focus on how to compute different roots precisely.
  • The \(n\)-th root of a number, \(\sqrt[n]{a}\), represents the number you multiply by itself \(n\) times to get \(a\).
  • Taking a root is essentially reversing an exponentiation process.
In the original exercise, finding the 5th root of \(\frac{1}{32}\) required recognizing \(\frac{1}{32}\) as \(2^{-5}\). Thus, \(\sqrt[5]{2^{-5}} = 2^{-1}\). Finally, applying the Negative Exponent Rule again, we simplified \(\frac{1}{2^{-1}}\) to get the answer \(2\). Recognizing these relationships makes evaluating expressions with roots straightforward and insightful.

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Most popular questions from this chapter

Find each root, if possible. $$\sqrt{5^{2}+12^{2}}$$

Rationalize the denominator of each expression. $$\frac{\sqrt{56}}{\sqrt{48}}$$

Rationalize the denominator and simplify completely. Assume the variables represent positive real numbers. $$\frac{10}{9-\sqrt{2}}$$

Rationalize the denominator of each expression. $$\frac{1}{\sqrt{5}}$$

What does it mean to rationalize the denominator of a radical expression?
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