Question

simplify. express your answer using positive exponents. \\(\frac{2q^{-1}}{q^{0}}\\)

Explanation:

Step1: Recall exponent rules

We know that \(a^0 = 1\) for any non - zero \(a\), and \(a^{-n}=\frac{1}{a^{n}}\) (where \(a
eq0\) and \(n\) is a positive integer). Also, when dividing with the same base \(a^{m}\div a^{n}=a^{m - n}\).

First, simplify the denominator \(q^{0}\). By the zero - exponent rule, \(q^{0}=1\) (assuming \(q
eq0\)). So our expression \(\frac{2q^{-1}}{q^{0}}\) becomes \(\frac{2q^{-1}}{1}=2q^{-1}\).

Step2: Convert negative exponent to positive

Using the negative - exponent rule \(a^{-n}=\frac{1}{a^{n}}\), we can rewrite \(q^{-1}\) as \(\frac{1}{q}\). So \(2q^{-1}=2\times\frac{1}{q}=\frac{2}{q}\).

Answer:

\(\frac{2}{q}\)