Exponential function negative: $\left(\frac{81}{4}\right)^{1/4}\left(\frac{1}4\right)^{-3/4}$

Question

This is another example. $\left(\dfrac{81}{4}\right)^{1/4}\left(\dfrac{1}4\right)^{-3/4}$

Multiply on both sides equals $\dfrac{81^{1/4}}{4^{1/4}}\cdot \dfrac{1^{-3/4}}{4^{-3/4}}$

This should be $\dfrac{3}{4^{1/4}}\cdot \dfrac{1^{-3/4}}{4^{-3/4}}$ which I get to -3/$\sqrt 4^{-1/2}$

Is that correct? I should get the answer 6.

Answer 1

Think I got the answer now. $$\frac{3}{2^{-1}} = 3 \cdot 2 = 6.$$

Answer 2

An alternative way (bring everything under a single exponent): $$(\frac{81}{4})^\frac 14 \cdot (\frac{1}{4})^\frac {-3}{4} = (\frac{81}{4})^\frac 14 \cdot [(\frac 14)^{-3}]^{\frac 14} = (\frac{81}{4})^\frac 14 \cdot(4^{3})^{\frac 14} = (\frac{81}{4}\cdot 64)^{\frac 14} = (81 \cdot 16)^{\frac 14} = 3 \cdot 2 = 6$$