Question

simplify the following expression to simplest form using only positive exponents.
\\((16x^{16}y^{8})^{\\ -\frac{1}{4}}\\)

Explanation:

Step1: Apply exponent rule \((ab)^n = a^n b^n\)

\((16x^{16}y^{8})^{-\frac{1}{4}} = 16^{-\frac{1}{4}} \cdot (x^{16})^{-\frac{1}{4}} \cdot (y^{8})^{-\frac{1}{4}}\)

Step2: Simplify \(16^{-\frac{1}{4}}\)

Since \(16 = 2^4\), \(16^{-\frac{1}{4}}=(2^4)^{-\frac{1}{4}} = 2^{4\times(-\frac{1}{4})}=2^{-1}=\frac{1}{2}\)

Step3: Simplify \((x^{16})^{-\frac{1}{4}}\)

Using \((a^m)^n = a^{mn}\), we get \(x^{16\times(-\frac{1}{4})}=x^{-4}=\frac{1}{x^4}\)

Step4: Simplify \((y^{8})^{-\frac{1}{4}}\)

Using \((a^m)^n = a^{mn}\), we get \(y^{8\times(-\frac{1}{4})}=y^{-2}=\frac{1}{y^2}\)

Step5: Multiply the results

\(16^{-\frac{1}{4}} \cdot (x^{16})^{-\frac{1}{4}} \cdot (y^{8})^{-\frac{1}{4}}=\frac{1}{2}\cdot\frac{1}{x^4}\cdot\frac{1}{y^2}=\frac{1}{2x^4y^2}\)

Answer:

\(\frac{1}{2x^4y^2}\)