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 \((a^m)^n=a^{m\times n}\), 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 rule for exponents

The product rule for exponents 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: Evaluate \(2^0\)

Any non - zero number raised to the power of 0 is 1, so \(2^0 = 1\).

Answer:

1