Answered

Select the expression that is equivalent to the expression below.

$16^{\frac{4}{9}}$


A$\sqrt[9]{4^4}$
B$\sqrt[9]{16^4}$
C$\sqrt[4]{16^9}$
D$\sqrt[4]{4^9}$

Verified Answer :

  • The expression \(16^{\frac{4}{9}}\) uses the identity \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\).
  • Substituting \(a=16, m=4, n=9\) yields \(\sqrt[9]{16^4}\).
  • Comparing this with the given choices confirms the correct expression is \(\sqrt[9]{16^4}\).
  • The final answer is \(\boxed{\sqrt[9]{16^4}}\).

Explanation

  1. Understanding the Exponential Property

To determine the expression equivalent to \(16^{\frac{4}{9}}\), we use the fundamental property of exponents and radicals. Any expression of the form \(a^{\frac{m}{n}}\) can be written as the \(n\)-th root of \(a\) raised to the power of \(m\), or expressed mathematically as \(a^{\frac{m}{n}} = \sqrt[n]{a^m}\).

  1. Applying the Property to the Given Expression

Applying this rule to our specific expression \(16^{\frac{4}{9}}\), we identify the base \(a = 16\), the numerator \(m = 4\), and the denominator \(n = 9\). By substituting these values into the formula \(\sqrt[n]{a^m}\), we obtain:
\[16^{\frac{4}{9}} = \sqrt[9]{16^4}\]

  1. Comparing with Options and Selecting the Correct Answer

We now compare this result, \(\sqrt[9]{16^4}\), with the provided options:

  1. \(\sqrt[9]{4^4}\)
  2. \(\sqrt[9]{16^4}\)
  3. \(\sqrt[4]{16^9}\)
  4. \(\sqrt[4]{4^9}\)

It is clear that the second option, \(\sqrt[9]{16^4}\), matches our derived expression perfectly. Thus, this is the correct equivalent form.

Examples

Understanding how to convert between exponential and radical forms is crucial in fields like physics and engineering, where complex formulas often appear with fractional exponents. For example, when calculating the power or energy in a system, you might encounter expressions like \(V^{\frac{2}{3}}\) or \(I^{\frac{1}{2}}\). Knowing how to rewrite these as \(\sqrt[3]{V^2}\) or \(\sqrt{I}\) allows you to use standard calculators or algebraic manipulation techniques to solve for variables efficiently, ensuring you can accurately model real-world phenomena.