Question

which expression is equivalent to \\(\frac{6^{-1}}{2^{-1}}\\)? \\(3^{-3}\cdot 3^{2}\\) \\(\frac{3^{-6}}{3^{6}}\\) \\(\frac{1}{3^{-1}}\\) \\(3^{0}\\)

Explanation:

Step1: Simplify the given expression

Recall the negative exponent rule: \(a^{-n}=\frac{1}{a^{n}}\) and \(\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}\). So, \(\frac{6^{-1}}{2^{-1}}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1}{6}\times\frac{2}{1}=\frac{2}{6}=\frac{1}{3}\)? Wait, no, wait: Wait, \(\frac{6^{-1}}{2^{-1}}=(6\times2^{-1})^{-1}\)? No, better to use the rule \(\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}\). So \(\frac{6^{-1}}{2^{-1}}=\frac{2^{1}}{6^{1}}=\frac{2}{6}=\frac{1}{3}\)? Wait, no, \(6^{-1}=\frac{1}{6}\), \(2^{-1}=\frac{1}{2}\), so dividing them: \(\frac{1/6}{1/2}=\frac{1}{6}\times2=\frac{2}{6}=\frac{1}{3}\). Wait, but also, \(6 = 2\times3\), so \(\frac{6^{-1}}{2^{-1}}=\frac{(2\times3)^{-1}}{2^{-1}}=\frac{2^{-1}\times3^{-1}}{2^{-1}} = 3^{-1}\)? Wait, no, \((ab)^{-1}=a^{-1}b^{-1}\), so \((2\times3)^{-1}=2^{-1}\times3^{-1}\), then \(\frac{2^{-1}\times3^{-1}}{2^{-1}} = 3^{-1}\)? Wait, that's not right. Wait, \(\frac{2^{-1}\times3^{-1}}{2^{-1}}=3^{-1}\), because \(2^{-1}\) cancels. But \(3^{-1}=\frac{1}{3}\). Now let's check each option:

Option 1: \(3^{-3}\cdot3^{2}\). Using exponent rule \(a^{m}\cdot a^{n}=a^{m + n}\), so \(3^{-3 + 2}=3^{-1}=\frac{1}{3}\). Wait, that's the same as the original expression. Wait, maybe my first calculation was wrong. Let's re - calculate the original expression:

\(\frac{6^{-1}}{2^{-1}}\). Using the rule \(\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}\), so \(\frac{6^{-1}}{2^{-1}}=\frac{2^{1}}{6^{1}}=\frac{2}{6}=\frac{1}{3}\). And \(3^{-3}\cdot3^{2}=3^{-3 + 2}=3^{-1}=\frac{1}{3}\). Let's check other options:

Option 2: \(\frac{3^{-6}}{3^{6}}\). Using \(a^{m}/a^{n}=a^{m - n}\), so \(3^{-6-6}=3^{-12}
eq\frac{1}{3}\).

Option 3: \(\frac{1}{3^{-1}}\). \(3^{-1}=\frac{1}{3}\), so \(\frac{1}{1/3}=3
eq\frac{1}{3}\).

Option 4: \(3^{0}=1
eq\frac{1}{3}\).

So the correct option is the first one: \(3^{-3}\cdot3^{2}\)

Step1: Simplify the given expression

We know that for any non - zero number \(a\) and integers \(m,n\), \(\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}\) and \(a^{m}\cdot a^{n}=a^{m + n}\).

First, simplify \(\frac{6^{-1}}{2^{-1}}\):

Using the rule \(\frac{a^{-m}}{b^{-n}}=\frac{b^{n}}{a^{m}}\), we have \(\frac{6^{-1}}{2^{-1}}=\frac{2^{1}}{6^{1}}=\frac{2}{6}=\frac{1}{3}\).

Step2: Simplify each option

• Option 1: \(3^{-3}\cdot3^{2}\)

Using the exponent rule \(a^{m}\cdot a^{n}=a^{m + n}\), we get \(3^{-3 + 2}=3^{-1}=\frac{1}{3}\).

• Option 2: \(\frac{3^{-6}}{3^{6}}\)

Using the exponent rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\), we get \(3^{-6-6}=3^{-12}
eq\frac{1}{3}\).

• Option 3: \(\frac{1}{3^{-1}}\)

Since \(3^{-1}=\frac{1}{3}\), then \(\frac{1}{3^{-1}} = 3
eq\frac{1}{3}\).

• Option 4: \(3^{0}\)

Any non - zero number to the power of 0 is 1, so \(3^{0}=1
eq\frac{1}{3}\).

Answer:

\(3^{-3}\cdot3^{2}\) (the first option)