Question

(\frac{3b}{(4b^{-4})^{4}})

Explanation:

Step1: Simplify the denominator using power of a product and power of a power rules.

The denominator is \((4b^{-4})^4\). By the power of a product rule \((ab)^n = a^n b^n\) and power of a power rule \((a^m)^n = a^{mn}\), we have:
\((4b^{-4})^4 = 4^4 \cdot (b^{-4})^4 = 256 \cdot b^{-16}\)
Since \(b^{-n}=\frac{1}{b^n}\), \(b^{-16}=\frac{1}{b^{16}}\), so the denominator becomes \(\frac{256}{b^{16}}\).

Step2: Rewrite the fraction as a multiplication.

The original expression \(\frac{3b}{(4b^{-4})^4}\) now becomes \(3b\div\frac{256}{b^{16}}\). Dividing by a fraction is multiplying by its reciprocal, so:
\(3b\times\frac{b^{16}}{256}\)

Step3: Multiply the terms with \(b\).

Using the rule \(a^m\times a^n = a^{m + n}\) for the \(b\) terms: \(b\times b^{16}=b^{1 + 16}=b^{17}\).
So the expression simplifies to \(\frac{3b^{17}}{256}\)

Answer:

\(\frac{3b^{17}}{256}\)