How do you simplify #(root3(6)*root4(6))^12#?

1 Answer
Mar 19, 2017

Answer:

#(root3(6)*root(4)6)^12=6^7=279936#

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 #color(red)star#, we see that:

#(root3(6)*root(4)6)^12=(6^(1/3)*6^(1/4))^12#

Now using #color(green)star#, this becomes:

#(6^(1/3)*6^(1/4))^12=(6^(1/3+1/4))^12#

Note that #1/3+1/4=4/12+3/12=7/12#.

#(6^(1/3+1/4))^12=(6^(7/12))^12#

Now using #color(blue)star#, we multiply the exponents:

#(6^(7/12))^12=6^(7/12xx12)#

And we see that #7/12xx12=7#:

#6^(7/12xx12)=6^7#

All of which we did without a calculator! For an expanded value, we could plug in #6^7# into a calculator to see that #6^7=279936#.