Question

simplify. express your answer using exponents.
$(g^{-2}h^{4})^{0}$

Explanation:

Step1: Recall the zero - exponent rule

The zero - exponent rule states that for any non - zero number \(x\), \(x^{0}=1\). Also, if we have a product of terms raised to a power, \((xy)^{n}=x^{n}y^{n}\), and when \(n = 0\), \((xy)^{0}=x^{0}y^{0}\) as long as \(x\) and \(y\) are non - zero.

In the expression \((g^{-2}h^{4})^{0}\), we can apply the power of a product rule first. According to the power of a product rule \((ab)^{n}=a^{n}b^{n}\), so \((g^{-2}h^{4})^{0}=(g^{-2})^{0}(h^{4})^{0}\).

Step2: Apply the zero - exponent rule to each term

For any non - zero base \(a\) and exponent \(m\), \((a^{m})^{n}=a^{mn}\). When \(n = 0\), \((a^{m})^{0}=a^{m\times0}=a^{0}\). And by the zero - exponent rule, \(a^{0}=1\) (assuming \(a
eq0\)).

For \((g^{-2})^{0}\), using the power of a power rule \((a^{m})^{n}=a^{mn}\), we have \(m=- 2\) and \(n = 0\), so \((g^{-2})^{0}=g^{-2\times0}=g^{0}\). By the zero - exponent rule, \(g^{0}=1\) (assuming \(g
eq0\)).

For \((h^{4})^{0}\), using the power of a power rule, we have \(m = 4\) and \(n=0\), so \((h^{4})^{0}=h^{4\times0}=h^{0}\). By the zero - exponent rule, \(h^{0}=1\) (assuming \(h
eq0\)).

Then \((g^{-2})^{0}(h^{4})^{0}=g^{0}h^{0}=1\times1 = 1\).

Alternatively, we can directly apply the rule that any non - zero expression raised to the power of \(0\) is \(1\). Since \(g^{-2}=\frac{1}{g^{2}}\) (for \(g
eq0\)) and \(h^{4}\) is a non - zero expression (for \(h
eq0\)), the product \(g^{-2}h^{4}\) is non - zero (as long as \(g
eq0\) and \(h
eq0\)), so \((g^{-2}h^{4})^{0}=1\).

Answer:

\(1\)