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Chapter 3: Problem 31

In Exercises 28鈥�35, find the indicated roots without using a calculator. $$ \sqrt[6]{64} $$

Short Answer

Expert verified
The sixth root of 64 is 2.

Step by step solution

01

Understand the Problem

We need to find the sixth root of 64, which is written as \(\sqrt[6]{64}\). This means we are looking for a number that, when multiplied by itself six times, equals 64.
02

Express 64 as a Power of 2

Recognize that 64 is a power of 2. Specifically, 64 can be written as \(64 = 2^6\).
03

Rewrite the Root Expression

Rewrite the original expression using the exponent form we found: \(\sqrt[6]{64} \). Since 64 is \(2^6\), we now have \( \sqrt[6]{2^6} \).
04

Apply Root and Exponent Rules

Use the properties of exponents to simplify the expression. The sixth root of \(2^6\) is \(2^{6/6} = 2^1 = 2\).
05

Verify the Solution

Verify the result by raising 2 to the sixth power: \(2^6 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\), which confirms our solution.

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Key Concepts

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

Exponents
Exponents are a way to express repeated multiplication of the same number. For instance, when we say '2 to the power of 6' (written as \(2^6\)), we mean multiplying 2 by itself six times:
  • \(2 \times 2 \times 2 \times 2 \times 2 \times 2\)

This helps simplify expressions and calculations. An important property of exponents is how they interact with roots: when you take the \(n\)th root of a number, you are essentially raising that number to the \(1/n\) exponent. For example, \(\sqrt[6]{64}\) is the same as \(64^{1/6}\). Understanding this helps us simplify and solve problems involving roots and exponents efficiently.
Root Properties
Root properties simplify the process of finding roots, especially when dealing with exponents. For example, the \(n\)th root of \(a^m\) is \(a^{m/n}\). This means the sixth root of \(64\repeatls\texttt{\tm}\). This relationship is powerful for solving root problems:
  • \(\sqrt[6]{64} = \sqrt[6]{2^6} = 2^{6/6} = 2^1 = 2\)

Another beneficial property is that the product of roots equals the root of the products:
  • \(\sqrt[n]{a} \times \sqrt[n]{b} = \sqrt[n]{a \times b}\)

Understanding these properties can help in breaking down complex root problems into simpler components.
Powers of 2
Understanding powers of 2 can make solving problems like the sixth root of 64 much simpler. Powers of 2 grow exponentially:
  • \(2^1 = 2\)
  • \(2^2 = 4\)
  • \(2^3 = 8\)
  • \(2^4 = 16\)
  • \(2^5 = 32\)
  • \(2^6 = 64\)

As you can see, 64 is a power of 2, specifically \(2^6\). This helps us understand that to find the sixth root of 64, we look for a base that, when raised to the power of 6, equals 64. Recognizing these patterns can greatly facilitate solving power and root problems efficiently.

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Most popular questions from this chapter

Rewrite each expression using a single base and a single exponent. $$2^{7} \cdot 2^{-4} \cdot 2^{x}$$

Write each fraction in lowest terms. $$ \frac{140}{196} $$

State whether the data in each table could be linear, and tell how you know. $$\begin{array}{|c|c|c|c|c|c|c|}\hline a & {-4} & {-3} & {-2} & {-1} & {0} & {1} \\ \hline b & {-32} & {-13.5} & {-4} & {-0.5} & {0} & {0.5} \\\ \hline\end{array}$$

Simplify each radical expression. If it is already simplified, say so. $$ \sqrt{17}-\sqrt{30} $$

According to the 1790 census, the population of the United States in 1790 was \(3,929,214 .\) You can approximate this value with powers of various numbers; for example, \(2^{21}\) is \(2,097,152\) and \(2^{22}\) is \(4,194,304\) . Using powers of \(2,\) the number \(2^{22}\) is the closest possible approximation to \(3,929,214\) . What is the closest possible approximation using powers of 3? Powers of 4? Powers of 5?
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