Problem:
(2^9/2^8 * sixth root of 3)^4
We have to solve it labeling each step.
I have it down to
(2^1 * sixth root of 3)^4 by the quotient of powers
I am stuck from there, though.
(2^9/2^8 * sixth root of 3)^4
You are getting closer.
2^9/2^8 = 2^1 or more commonly known as 2.
because 2^(9-8) = 2^1
sixth root of three can be written as 3^(1/6)
so.
[ 2*3^(1/6) ]^4
Apply the 4 to inside the bracket
2^4 * 3^[ (1/6)*4) ]
16 * 3^(4/6)
16 * 3^(2/3)
=33.281341
Hey there lmschneider,
It's a bit hard to establish what your question looked like from the notation, but taking it literally...
(2^9/2^8 * sixth root of 3)^4 is $\displaystyle (\frac {2^9}{2^8} * \sqrt[6]{3}) ^4$
If this is the case...
$\displaystyle (\frac {2^9}{2^8} * \sqrt[6]{3}) ^4 = (2 \sqrt[6]{3})^4$
$\displaystyle = (2 * 3^{\frac{1}{6}})^4$
$\displaystyle = (2^4 * 3^{\frac{4}{6}})$
$\displaystyle = 2^4 * 3^{\frac{2}{3}}$
$\displaystyle = 16 * \sqrt[3]{3^2}$
$\displaystyle = 16 \sqrt[3]{9}$
Trust this helps...