Question

evaluate. write your answer as a whole number or as a simplified fraction.
$6^{-1} \cdot 3^{-4} = \frac{\square}{\square}$

Explanation:

Step1: Recall negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^{n}}\) for any non - zero number \(a\) and positive integer \(n\).
Applying this rule to \(6^{-1}\) and \(3^{-4}\), we get \(6^{-1}=\frac{1}{6^{1}}=\frac{1}{6}\) and \(3^{-4}=\frac{1}{3^{4}}\).
Since \(3^{4}=3\times3\times3\times3 = 81\), then \(3^{-4}=\frac{1}{81}\).

Step2: Multiply the two fractions

We need to find the product of \(\frac{1}{6}\) and \(\frac{1}{81}\). When multiplying two fractions \(\frac{a}{b}\times\frac{c}{d}=\frac{a\times c}{b\times d}\), here \(a = 1\), \(b = 6\), \(c = 1\), \(d=81\).
So \(\frac{1}{6}\times\frac{1}{81}=\frac{1\times1}{6\times81}=\frac{1}{486}\).

Answer:

\(\frac{1}{486}\)