Question

express the following fraction in simplest form, only using positive exponents.
\frac{(2p^{-1})^{5}}{6j^{-8}p^{-8}}

Explanation:

Step1: Simplify the numerator using power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). For \((2p^{-1})^5\), we have:
\((2p^{-1})^5 = 2^5\times(p^{-1})^5 = 32p^{-5}\) (since \((a^m)^n=a^{mn}\), so \((p^{-1})^5 = p^{-5}\))

Step2: Rewrite the fraction with the simplified numerator

Now our fraction becomes \(\frac{32p^{-5}}{6j^{-8}p^{-8}}\)

Step3: Simplify the coefficients and use the quotient rule for exponents

The quotient rule for exponents is \(\frac{a^m}{a^n}=a^{m - n}\). First, simplify the coefficients: \(\frac{32}{6}=\frac{16}{3}\)
For the \(p\) terms: \(\frac{p^{-5}}{p^{-8}}=p^{-5-(-8)} = p^{-5 + 8}=p^{3}\)
For the \(j\) term: \(\frac{1}{j^{-8}}=j^{8}\) (since \(\frac{1}{a^{-n}}=a^{n}\))

Step4: Combine the simplified terms

Multiply the simplified coefficient, \(p\) term, and \(j\) term together: \(\frac{16}{3}\times p^{3}\times j^{8}=\frac{16j^{8}p^{3}}{3}\)

Answer:

\(\frac{16j^{8}p^{3}}{3}\)