How do you simplify #(root3(6)*root4(6))^12#?
1 Answer
Mar 19, 2017
Explanation:
We need the rules:
#rootmx=x^(1/m)" "" "color(red)star# #x^a*x^b=x^(a+b)" "" "color(green)star# #(x^c)^d=x^(cd)" "" "color(blue)star#
Using
#(root3(6)*root(4)6)^12=(6^(1/3)*6^(1/4))^12#
Now using
#(6^(1/3)*6^(1/4))^12=(6^(1/3+1/4))^12#
Note that
#(6^(1/3+1/4))^12=(6^(7/12))^12#
Now using
#(6^(7/12))^12=6^(7/12xx12)#
And we see that
#6^(7/12xx12)=6^7#
All of which we did without a calculator! For an expanded value, we could plug in