Question

express the following fraction in simplest form, only using positive exp\\(\frac{6k}{(3k^{-4})^2}\\)

Explanation:

Step1: Expand the denominator

First, apply the exponent rule $(ab)^n = a^n b^n$ and $(x^m)^n = x^{mn}$ to the denominator:
$$(3k^{-4})^2 = 3^2 \cdot (k^{-4})^2 = 9k^{-8}$$
The expression becomes:
$$\frac{6k}{9k^{-8}}$$

Step2: Simplify coefficients and exponents

Simplify the numerical fraction $\frac{6}{9} = \frac{2}{3}$. Then use the exponent rule $\frac{x^m}{x^n} = x^{m-n}$ for the variable terms:
$$k^{1 - (-8)} = k^{1+8} = k^9$$
Combine the results:
$$\frac{2}{3}k^9$$

Answer:

$\frac{2}{3}k^9$