Question

select all the expressions that are equivalent to $2^2 \cdot 6^2$. options: $\frac{1}{12^2}$, $12^2$, $\frac{1}{12^{-2}}$, $12^4$

Explanation:

Step1: Simplify the original expression

First, recall the exponent rule \(a^n\cdot b^n=(a\cdot b)^n\). For the expression \(2^{2}\cdot6^{2}\), we can apply this rule:
\(2^{2}\cdot6^{2}=(2\times6)^{2}\)
\(2\times6 = 12\), so \((2\times6)^{2}=12^{2}\)

Step2: Recall the negative exponent rule

The negative exponent rule states that \(a^{-n}=\frac{1}{a^{n}}\) (and \(\frac{1}{a^{-n}}=a^{n}\)). Let's rewrite \(12^{2}\) using negative exponents. We know that \(12^{2}=\frac{1}{12^{- 2}}\) (because if we let \(a = 12\), \(n=2\), then \(a^{n}=\frac{1}{a^{-n}}\))

Now let's analyze each option:

• \(\frac{1}{12^{2}}\): This is \(12^{-2}\), which is not equal to \(12^{2}\) (or \(2^{2}\cdot6^{2}\)), so this is incorrect.
• \(12^{2}\): We showed in Step 1 that \(2^{2}\cdot6^{2}=(2\times6)^{2}=12^{2}\), so this is correct.
• \(\frac{1}{12^{-2}}\): From the negative exponent rule, \(\frac{1}{12^{-2}}=12^{2}\) (since \(a^{-n}=\frac{1}{a^{n}}\) implies \(\frac{1}{a^{-n}}=a^{n}\)), so this is equivalent to \(12^{2}\) (and thus to \(2^{2}\cdot6^{2}\)), so this is correct.
• \(12^{4}\): \(12^{4}=12^{2 + 2}=12^{2}\times12^{2}\), which is not equal to \(12^{2}\) (unless \(12^{2}=1\), which it isn't), so this is incorrect.

Answer:

The equivalent expressions are \(12^{2}\) and \(\frac{1}{12^{-2}}\) (i.e., the options with \(12^{2}\) and \(\frac{1}{12^{-2}}\) should be selected).