Question

express the following fraction in simplest form, only using positive exponents. \\(\frac{-3(v^3)^{-5}}{6s^{-7}}\\).

Explanation:

Step1: Simplify the coefficient and use exponent rules for \(v\)

First, simplify the coefficient \(\frac{-3}{6}=-\frac{1}{2}\). For the exponent of \(v\), use the power of a power rule \((a^m)^n = a^{mn}\), so \((v^3)^{-5}=v^{3\times(-5)} = v^{-15}\). Now the expression becomes \(\frac{-\frac{1}{2}v^{-15}}{s^{-7}}\).

Step2: Use the negative exponent rule \(a^{-n}=\frac{1}{a^n}\)

Recall that \(a^{-n}=\frac{1}{a^n}\) and \(\frac{1}{a^{-n}} = a^n\). So \(v^{-15}=\frac{1}{v^{15}}\) and \(\frac{1}{s^{-7}}=s^{7}\). Substituting these in, we get \(-\frac{1}{2}\times\frac{1}{v^{15}}\times s^{7}=-\frac{s^{7}}{2v^{15}}\).

Answer:

\(\boxed{-\dfrac{s^{7}}{2v^{15}}}\)