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:
\(4(ya^{-5})^{-1}=4y^{-1}a^{(-5)\times(-1)} = 4y^{-1}a^{5}\)

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

The coefficient of numerator and denominator is 4, so \(\frac{4}{4} = 1\). 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 \(a^m\div a^n=a^{m - n}\)).
Combining these results: \(1\times y^{7}\times a=\frac{y^{7}a}{1}=y^{7}a\) (or \(a y^{7}\))

Answer:

\(a y^{7}\) (or \(y^{7}a\))