Question

(81)^{-\frac{3}{2}}

Explanation:

Step1: Recall exponent rules

We know that \(a^{-n}=\frac{1}{a^{n}}\) and \((a^{m})^{n}=a^{mn}\), also \(81 = 3^{4}\) or \(81=9^{2}\), we can use \(81 = 3^{4}\) here. So \((81)^{-\frac{3}{2}}=\frac{1}{(81)^{\frac{3}{2}}}\)

Step2: Simplify \((81)^{\frac{3}{2}}\)

First, \((81)^{\frac{3}{2}}=((81)^{\frac{1}{2}})^{3}\). And \((81)^{\frac{1}{2}}=\sqrt{81} = 9\) (since we take the positive square root for real - valued exponentiation in this context). Then \(((81)^{\frac{1}{2}})^{3}=9^{3}\)

Step3: Calculate \(9^{3}\)

\(9^{3}=9\times9\times9 = 729\)

Step4: Find the final value

Since \((81)^{-\frac{3}{2}}=\frac{1}{(81)^{\frac{3}{2}}}\) and \((81)^{\frac{3}{2}} = 729\), then \((81)^{-\frac{3}{2}}=\frac{1}{729}\)

Answer:

\(\frac{1}{729}\)