Question

express the following fraction in simplest form, only using positive exponents. \\(\frac{5a^{4}}{-2(a^{4}m^{-4})^{-1}}\\)

Explanation:

Step1: Apply the negative exponent rule

Recall that \((x^m)^n = x^{mn}\) and \((ab)^n = a^n b^n\), and \(x^{-n}=\frac{1}{x^n}\) or \(\frac{1}{x^{-n}} = x^n\). First, simplify the denominator \(-2(a^{4}m^{-4})^{-1}\). Using the power - of - a - product rule and the power - of - a - power rule: \((a^{4}m^{-4})^{-1}=a^{4\times(-1)}m^{-4\times(-1)} = a^{-4}m^{4}\). So the denominator becomes \(-2\times a^{-4}m^{4}=\frac{-2m^{4}}{a^{4}}\) (since \(a^{-4}=\frac{1}{a^{4}}\)). Now our fraction is \(\frac{5a^{4}}{\frac{-2m^{4}}{a^{4}}}\).

Step2: Divide by a fraction (multiply by its reciprocal)

Dividing by a fraction \(\frac{A}{B}\) is the same as multiplying by its reciprocal \(\frac{B}{A}\). So \(\frac{5a^{4}}{\frac{-2m^{4}}{a^{4}}}=5a^{4}\times\frac{a^{4}}{-2m^{4}}\).

Step3: Multiply the numerators and apply the product rule for exponents

Using the rule \(x^m\times x^n=x^{m + n}\), for the \(a\) terms: \(a^{4}\times a^{4}=a^{4 + 4}=a^{8}\). The numerator of the product is \(5\times a^{8}\) and the denominator is \(-2m^{4}\). So we get \(\frac{5a^{8}}{-2m^{4}}=-\frac{5a^{8}}{2m^{4}}\).

Answer:

\(-\frac{5a^{8}}{2m^{4}}\)