Question

2^{-1} cdot 2^{\frac{7}{4}}

Explanation:

Step1: Apply exponent rule \(a^m \cdot a^n = a^{m + n}\)

For \(2^{-1} \cdot 2^{\frac{7}{4}}\), we add the exponents: \(-1+\frac{7}{4}\)

Step2: Calculate the sum of exponents

\(-1+\frac{7}{4}=\frac{-4 + 7}{4}=\frac{3}{4}\)
So \(2^{-1} \cdot 2^{\frac{7}{4}}=2^{\frac{3}{4}}\)

Step3: Rewrite \(2^{\frac{3}{4}}\) as a radical (optional, but to simplify)

\(2^{\frac{3}{4}}=\sqrt[4]{2^3}=\sqrt[4]{8}\) or keep it as \(2^{\frac{3}{4}}\). If we want a numerical approximation, \(2^{\frac{3}{4}}\approx2^{0.75}\approx1.6818\) (but the exact form is \(2^{\frac{3}{4}}\) or \(\sqrt[4]{8}\))

Answer:

The exact value is \(2^{\frac{3}{4}}\) (or \(\sqrt[4]{8}\), approximately \(1.68\) if decimal is preferred)