Question

Evaluate the ff. equation: (2*sqrt of 2 multiplied to 2^1/4) all over 2^3/4. Step by step instructions, please?

Answer 1

Here are the tools:
\[3^\frac{1}{4}*3^b*3^2=3^{\frac{1}{4}+b+2}\]
\[\frac{5^a*5^b}{5^c}=5^{a+b-c}\]
\[\sqrt{7} = 7^\frac{1}{2}\]

Answer 2

What if the bases are different?

Answer 3

then you're doomed, except for maybe if powers are identical: \[\sqrt{6}=6^\frac{1}{2}=2^\frac{1}{2}*3^\frac{1}{2}\]

Answer 4

... That's very reassuring. :)) Thank you for the help!

Same goes for something like this?
\[\frac{6\sqrt6(2^{\frac{-1}{4}})}{3\sqrt6(2^{\frac{3}{4}})}\]

Answer 5

split the 2s and 3s and treat them separately. my final answer had 2^stuff times 3^(otherstuff)

Answer 6

Mm. Thank you, then. :) Still comprehending, though.

Answer 7

Cancellation properties give you\[\frac{6\sqrt6(2^{\frac{-1}{4}})}{3\sqrt6(2^{\frac{3}{4}})} = \frac{2(2^\frac{-1}{4})}{2^{\frac{3}{4}}}=2^{1-\frac{1}{4}-\frac{3}{4}}\]

Answer 8

Yeah, thank you very much. :D