Question

c. \\(4^2 \times 2^{-4}\\)

Explanation:

Step1: Rewrite 4 as a power of 2

Since \(4 = 2^2\), we can rewrite \(4^2\) as \((2^2)^2\). Using the power - of - a - power rule \((a^m)^n=a^{mn}\), we get \((2^2)^2 = 2^{2\times2}=2^4\). So the expression \(4^2\times2^{-4}\) becomes \(2^4\times2^{-4}\).

Step2: Use the product of powers rule

The product of powers rule states that \(a^m\times a^n=a^{m + n}\). For \(a = 2\), \(m = 4\) and \(n=-4\), we have \(2^4\times2^{-4}=2^{4+( - 4)}\).

Step3: Simplify the exponent

Simplify the exponent \(4+( - 4)=0\). So we have \(2^0\).

Step4: Use the zero - exponent rule

The zero - exponent rule states that \(a^0 = 1\) for \(a
eq0\). Since \(2
eq0\), \(2^0 = 1\).

Answer:

\(1\)