Question

question
express the following fraction in simplest form, only using positive exponents.
\\(\dfrac{-4(k^{-3}v^4)^{-1}}{20k^{-2}v^8}\\)

Explanation:

Step1: Simplify the coefficient and use exponent rule \((a^m)^n = a^{mn}\)

First, simplify the coefficient \(\frac{-4}{20}=-\frac{1}{5}\). Then, apply the exponent rule \((k^{-3}v^{4})^{-1}=k^{(-3)\times(-1)}v^{4\times(-1)} = k^{3}v^{-4}\). So the expression becomes \(\frac{-\frac{1}{5}k^{3}v^{-4}}{k^{-2}v^{8}}\).

Step2: Use exponent rule \(\frac{a^m}{a^n}=a^{m - n}\) for \(k\) and \(v\)

For the \(k\) terms: \(\frac{k^{3}}{k^{-2}}=k^{3-(-2)} = k^{5}\). For the \(v\) terms: \(\frac{v^{-4}}{v^{8}}=v^{-4 - 8}=v^{-12}=\frac{1}{v^{12}}\) (since we need positive exponents). Combine with the coefficient: \(-\frac{1}{5}\times k^{5}\times\frac{1}{v^{12}}=-\frac{k^{5}}{5v^{12}}\).

Answer:

\(-\frac{k^{5}}{5v^{12}}\)