Question

express the following fraction in simplest form, only using positive exponents.\\(\frac{5(k^{-1})^{3}}{-3k^{-2}}\\)

Explanation:

Step1: Simplify the exponent of \(k\) in the numerator

Using the power of a power rule \((a^m)^n = a^{mn}\), for \((k^{-1})^3\), we have \(k^{-1\times3}=k^{-3}\). So the numerator becomes \(5k^{-3}\).
The expression is now \(\frac{5k^{-3}}{-3k^{-2}}\).

Step2: Use the quotient rule for exponents

The quotient rule for exponents is \(\frac{a^m}{a^n}=a^{m - n}\). For the \(k\) terms, \(\frac{k^{-3}}{k^{-2}}=k^{-3-(-2)}=k^{-1}\). Now the expression is \(\frac{5k^{-1}}{-3}\).

Step3: Convert negative exponent to positive

Recall that \(a^{-n}=\frac{1}{a^n}\), so \(k^{-1}=\frac{1}{k}\). Substituting this in, we get \(\frac{5\times\frac{1}{k}}{-3}=\frac{5}{-3k}=-\frac{5}{3k}\).

Answer:

\(-\frac{5}{3k}\)