Question

question
which expression is equivalent to \\(\frac{3^{-2}}{(3^2)^{-1}}\\)?

answer
\\(\frac{1}{81}\\) \\(\frac{1}{3}\\)
1 0

Explanation:

Step1: Simplify the exponents in denominator

First, simplify \((3^2)^{-1}\). Using the power - of - a - power rule \((a^m)^n=a^{m\times n}\), we have \((3^2)^{-1}=3^{2\times(-1)} = 3^{-2}\). So the original expression \(\frac{3^{-2}}{(3^2)^{-1}}\) becomes \(\frac{3^{-2}}{3^{-2}}\).

Step2: Divide the two terms with the same base

When dividing two terms with the same base \(a\) (where \(a
eq0\)) and exponents \(m\) and \(n\), we use the rule \(\frac{a^m}{a^n}=a^{m - n}\). Here, \(a = 3\), \(m=-2\), and \(n = - 2\). So \(\frac{3^{-2}}{3^{-2}}=3^{-2-(-2)}\).

Step3: Simplify the exponent

Simplify the exponent: \(-2-(-2)=-2 + 2=0\). Then \(3^{0}\) is equal to \(1\) (since any non - zero number \(a\) raised to the power of \(0\) is \(1\), i.e., \(a^{0}=1,a
eq0\)).

Answer:

\(1\) (the option corresponding to \(1\))