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Chapter 10: Problem 27

Write in radical form and evaluate. \(\left(\frac{16}{81}\right)^{3 / 4}\)

Short Answer

Expert verified
The given expression \(\left(\frac{16}{81}\right)^{3 / 4}\) can be converted into a radical expression as \(\sqrt[4]{\left(\frac{16}{81}\right)^3}\). By applying the exponent to the numerator and denominator, we get \(\sqrt[4]{\frac{4096}{531441}}\), which simplifies to \(\frac{8}{27}\).

Step by step solution

01

Write the expression as a radical

To convert the expression \(\left(\frac{16}{81}\right)^{3/4}\) into a radical expression, we can rewrite the exponent as a root. The expression becomes: \[\sqrt[4]{\left(\frac{16}{81}\right)^3}\]
02

Apply the exponent to the numerator and denominator

Now, apply the power of 3 (from the exponent) to both the numerator and the denominator inside the parenthesis. We get: \[\sqrt[4]{\frac{16^3}{81^3}}\]
03

Calculate the powers for the numerator and denominator

We now need to compute the cube of both 16 and 81. Thus, we have: \[16^3 = (16)(16)(16) = 4096\] \[81^3 = (81)(81)(81) = 531441\] Now, substitute these values back into the expression: \[\sqrt[4]{\frac{4096}{531441}}\]
04

Evaluate the fourth root

Finally, we need to calculate the fourth root of the fraction, which means finding a number that multiplied by itself four times equals the fraction. We know that \((2^4 = 16)\) and \((3^4 = 81)\). By applying this information and properties of exponents, we can write the expression as: \[\sqrt[4]{\frac{2^3 \times 2^3 \times 2^3}{3^3 \times 3^3 \times 3^3 脳 3^3}} = \frac{2^3}{3^3} = \frac{8}{27}\] Therefore, in radical form, we have that \(\left(\frac{16}{81}\right)^{3 / 4} = \frac{8}{27}\).

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Key Concepts

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

Exponentiation
Exponentiation is a mathematical operation involving numbers called bases and exponents. Generally, it is written as \(a^n\), where \(a\) is the base, and \(n\) is the exponent. This simply means that the base, \(a\), is multiplied by itself \(n\) times. For example:
  • \(2^3\) equals \(2 \times 2 \times 2 = 8\)
  • \(( rac{16}{81})^3\) means \(( rac{16}{81})\) is multiplied by itself three times.
Exponentiation is a fundamental concept in algebra because it helps in simplifying expressions and solving equations. When dealing with fractions, the exponent is applied to both the numerator and the denominator separately.
Roots
Roots are the inverse operation of exponentiation. A root asks the question, "What number, when raised to a certain power, gives the original number?" The most common is the square root, but other roots like the cube root and the fourth root are also used. In this exercise,
  • The fourth root of \(\left(\frac{16}{81}\right)\) is simplified from the expression \(\left(\frac{16}{81}\right)^{3/4}\).
  • This was done by converting the fractional exponent \(3/4\) into a radical expression.
Finding roots is crucial in solving equations and simplifying radical expressions, as it helps reduce complex expressions into more manageable forms.
Fractional Exponents
Fractional exponents are a handy way to express roots and powers together in one expression. For instance, \(a^{m/n}\) can be written in radical form as \(\sqrt[n]{a^m}\). This provides an alternative view to understand and simplify expressions like \((\frac{16}{81})^{3/4}\):
  • The numerator "3" of the exponent indicates that the expression should be raised to the third power.
  • The denominator "4" suggests extracting the fourth root of the resulting expression.
Thus, it can be seen as first taking the fourth root, then cubing the result, simplifying the mathematical process.
Algebraic Manipulation
Algebraic manipulation involves transforming and simplifying algebraic expressions using various algebraic rules. This technique is applied in the solution to the exercise, where step-by-step simplifications guide the conversion from a complex expression to a simpler form.
  • By breaking down \((\frac{16}{81})^{3/4}\) into \(\sqrt[4]{\frac{4096}{531441}}\), we apply the concept of distributing exponentiation over division.
  • Moreover, the exercise leverages the property of distributing roots over products to individually simplify the numerator \(16^3\) and the denominator \(81^3\).
Such operations are fundamental in algebra, allowing students to confidently handle more complex equations and expressions by understanding how to rearrange and simplify them effectively.

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