Question

express the following fraction in simplest form, only using positive exponents.
\\(\frac{4(ya^{-5})^{-1}}{4y^{-8}a^4}\\)

Explanation:

Step1: Simplify the numerator's exponent

Using the power of a product rule \((ab)^n = a^n b^n\) and the power of a power rule \((a^m)^n = a^{mn}\), we have \((ya^{-5})^{-1}=y^{-1}a^{(-5)\times(-1)} = y^{-1}a^{5}\). So the numerator becomes \(4y^{-1}a^{5}\).

Step2: Simplify the fraction by dividing coefficients and using exponent rules for like bases

The coefficient is \(\frac{4}{4}=1\). For the \(y\) terms: \(y^{-1}\div y^{-8}=y^{-1 - (-8)} = y^{7}\) (using \(a^m\div a^n=a^{m - n}\)). For the \(a\) terms: \(a^{5}\div a^{4}=a^{5 - 4}=a^{1}=a\) (using \(a^m\div a^n=a^{m - n}\)).

Step3: Combine the results

Multiplying the coefficient, \(y\) term, and \(a\) term together, we get \(1\times y^{7}\times a = y^{7}a\).

Answer:

\(y^{7}a\)