Question

express the following fraction in simplest form, only using positive exponents. \\(\frac{(3d^{-4}n^{5})^{-3}}{12d^{3}n^{4}}\\)

Explanation:

Step1: Apply power of a product rule

Using \((ab)^n = a^n b^n\) and \((a^m)^n = a^{mn}\) on the numerator \((3d^{-4}n^{5})^{-3}\):
\(3^{-3}(d^{-4})^{-3}(n^{5})^{-3}=3^{-3}d^{12}n^{-15}\)
So the expression becomes \(\frac{3^{-3}d^{12}n^{-15}}{12d^{3}n^{4}}\)

Step2: Simplify the coefficient and use exponent rules for division

First, simplify the coefficient: \(3^{-3}=\frac{1}{27}\), and \(12 = 3\times4\), so \(\frac{3^{-3}}{12}=\frac{1}{27\times12}=\frac{1}{324}\)
For the \(d\) terms: \(d^{12}\div d^{3}=d^{12 - 3}=d^{9}\) (using \(a^m\div a^n=a^{m - n}\))
For the \(n\) terms: \(n^{-15}\div n^{4}=n^{-15 - 4}=n^{-19}=\frac{1}{n^{19}}\) (using \(a^m\div a^n=a^{m - n}\) and \(a^{-n}=\frac{1}{a^n}\))

Step3: Combine all parts

Putting it together: \(\frac{1}{324}\times d^{9}\times\frac{1}{n^{19}}=\frac{d^{9}}{324n^{19}}\)
We can also simplify \(\frac{1}{324}=\frac{1}{4\times81}=\frac{1}{4\times9^{2}}\), but \(\frac{d^{9}}{324n^{19}}\) is in simplest form with positive exponents.

Answer:

\(\frac{d^{9}}{324n^{19}}\)