Question

question
which expression is equivalent to \\(\frac{(4^0)^{-3}}{4^{-3}}\\)?
answer
\\(\circ\\) 0 \\(\circ\\) 64
\\(\circ\\) 16 \\(\circ\\) 1

Explanation:

Step1: Simplify the numerator's exponent

Recall that any non - zero number to the power of 0 is 1, so \(4^{0}=1\). Then we have \((4^{0})^{-3}=1^{-3}\). And any non - zero number to the power of - n is \(\frac{1}{n}\), but for \(a = 1\), \(1^{-3}=\frac{1}{1^{3}} = 1\) (since \(1\) to any power is \(1\)).

Step2: Simplify the denominator

The denominator is \(4^{-3}\). Using the rule \(a^{-n}=\frac{1}{a^{n}}\), we can rewrite \(4^{-3}\) as \(\frac{1}{4^{3}}\). And \(4^{3}=4\times4\times4 = 64\), so \(4^{-3}=\frac{1}{64}\).

Step3: Divide the numerator by the denominator

We have \(\frac{(4^{0})^{-3}}{4^{-3}}=\frac{1}{\frac{1}{64}}\). Dividing by a fraction is the same as multiplying by its reciprocal, so \(\frac{1}{\frac{1}{64}}=1\times64 = 64\).

Answer:

64