Question

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

Explanation:

Step1: Apply the power of a product rule to the numerator

The power of a product rule states that \((ab)^n = a^n b^n\). So for \((2p^{-1})^5\), we have \(2^5\times(p^{-1})^5\). Calculating \(2^5 = 32\) and using the power of a power rule \((a^m)^n=a^{mn}\), we get \((p^{-1})^5 = p^{-5}\). So the numerator becomes \(32p^{-5}\).

Step2: Rewrite the fraction with the new numerator

The fraction is now \(\frac{32p^{-5}}{6j^{-8}p^{-8}}\). We can split this into the product of three fractions: \(\frac{32}{6}\times\frac{p^{-5}}{p^{-8}}\times\frac{1}{j^{-8}}\).

Step3: Simplify the coefficient fraction

Simplify \(\frac{32}{6}\) by dividing numerator and denominator by 2, we get \(\frac{16}{3}\).

Step4: Simplify the \(p\)-term using the quotient rule for exponents

The quotient rule for exponents is \(\frac{a^m}{a^n}=a^{m - n}\). For \(\frac{p^{-5}}{p^{-8}}\), we have \(p^{-5-(-8)}=p^{-5 + 8}=p^{3}\).

Step5: Simplify the \(j\)-term using the negative exponent rule

The negative exponent rule is \(a^{-n}=\frac{1}{a^{n}}\) or \(\frac{1}{a^{-n}}=a^{n}\). So \(\frac{1}{j^{-8}}=j^{8}\).

Step6: Multiply all the simplified parts together

Multiplying \(\frac{16}{3}\times p^{3}\times j^{8}\), we get \(\frac{16j^{8}p^{3}}{3}\).

Answer:

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