Question

which radical expression is equivalent to ( k^{-\frac{2}{9}} )?
choose 1 answer
a ( sqrt3{k} )
b ( \frac{1}{(sqrt{k})^{9}} )
c ( \frac{1}{sqrt9{k^{2}}} )
d ( sqrt9{k^{2}} )

Explanation:

Step1: Recall the exponent - radical rule

The rule for converting a rational exponent to a radical is \(a^{\frac{m}{n}}=\sqrt[n]{a^{m}}\) when \(n>0\), and also \(a^{-\frac{m}{n}}=\frac{1}{a^{\frac{m}{n}}}=\frac{1}{\sqrt[n]{a^{m}}}\). Here, we have \(k^{-\frac{3}{9}}\), first simplify the exponent \(-\frac{3}{9}=-\frac{1}{3}\). So we need to find the radical equivalent of \(k^{-\frac{1}{3}}\).

Step2: Apply the negative exponent rule

Using the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\), for \(a = k\) and \(n=\frac{1}{3}\), we have \(k^{-\frac{1}{3}}=\frac{1}{k^{\frac{1}{3}}}\).

Step3: Convert the positive rational exponent to radical

Using the rule \(a^{\frac{1}{n}}=\sqrt[n]{a}\), for \(a = k\) and \(n = 3\), we get \(k^{\frac{1}{3}}=\sqrt[3]{k}\). So \(\frac{1}{k^{\frac{1}{3}}}=\frac{1}{\sqrt[3]{k}}\).

Answer:

\(\frac{1}{\sqrt[3]{k}}\) (corresponding to the option with \(\frac{1}{\sqrt[3]{k}}\))