Question

question
which expression is equivalent to (left(2^{-1}cdot 2^{3}
ight)^{-1})?
answer
(\bigcirc) 16 (\bigcirc) (\frac{1}{16})
(\bigcirc) (\frac{1}{2}) (\bigcirc) (\frac{1}{4})

Explanation:

Step1: Simplify the exponent inside the parentheses

When multiplying exponents with the same base, we add the exponents. So for \(2^{-1}\cdot2^{3}\), we use the rule \(a^{m}\cdot a^{n}=a^{m + n}\). Here, \(a = 2\), \(m=-1\), and \(n = 3\). So \(2^{-1+3}=2^{2}\).

Step2: Simplify the outer exponent

Now we have \((2^{2})^{-1}\). When raising a power to a power, we multiply the exponents. Using the rule \((a^{m})^{n}=a^{m\times n}\), with \(a = 2\), \(m = 2\), and \(n=-1\), we get \(2^{2\times(-1)}=2^{-2}\).

Step3: Convert negative exponent to positive

A negative exponent means the reciprocal. So \(2^{-2}=\frac{1}{2^{2}}\). And \(2^{2}=4\), so \(\frac{1}{2^{2}}=\frac{1}{4}\)? Wait, no, wait. Wait, let's check again. Wait, step 1: \(2^{-1}\cdot2^{3}=2^{-1 + 3}=2^{2}=4\). Then step 2: \((4)^{-1}=\frac{1}{4}\)? Wait, no, wait, the original expression is \((2^{-1}\cdot2^{3})^{-1}\). Wait, maybe I made a mistake. Let's redo step 1: \(2^{-1}\cdot2^{3}=2^{-1 + 3}=2^{2}=4\). Then \((4)^{-1}=\frac{1}{4}\)? Wait, no, wait, \(2^{2}=4\), so \((4)^{-1}=\frac{1}{4}\)? But wait, let's do it with exponents again. Wait, \(2^{-1}\cdot2^{3}=2^{2}\), then \((2^{2})^{-1}=2^{-2}=\frac{1}{2^{2}}=\frac{1}{4}\)? Wait, but let's compute the original expression numerically. \(2^{-1}=\frac{1}{2}\), \(2^{3}=8\), so \(\frac{1}{2}\times8 = 4\). Then \((4)^{-1}=\frac{1}{4}\). So the equivalent expression is \(\frac{1}{4}\). Wait, but the options are \(\frac{1}{16}\), \(\frac{1}{4}\), etc. Wait, maybe I made a mistake in step 1. Wait, \(2^{-1}\cdot2^{3}\): exponent addition is correct, \(-1 + 3 = 2\), so \(2^{2}=4\). Then \((4)^{-1}=\frac{1}{4}\). So the answer is \(\frac{1}{4}\)? Wait, but let's check the options. The options are 16, \(\frac{1}{16}\), \(\frac{1}{2}\), \(\frac{1}{4}\). So \(\frac{1}{4}\) is one of the options. Wait, but let's do it again. Wait, \(2^{-1}\cdot2^{3}=2^{-1 + 3}=2^{2}\), then \((2^{2})^{-1}=2^{-2}=\frac{1}{2^{2}}=\frac{1}{4}\). Yes, that's correct.

Answer:

\(\frac{1}{4}\) (the option corresponding to \(\frac{1}{4}\))