How do you simplify #(27 ^(1/12) * 27 ^(-5/12))^-2#?

Question

1597 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-21T01:42:33

#9#

Keep in mind that

#(a^b)^c = a^(b*c)#

and

#a^b*a^c = a^(b+c)#

and

#a^(b/c)=root(c)(a^b)#

and

#root(b)(a^b) = a#

`````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````````

Use the first concept above and apply it to the first step.

#(27^(1/12)*27^(-5/12))^-2#

#(27^(1/12*-2/1) * 27^(-5/12*-2/1))#

#(27^(-2/12) * 27^(10/12))#

Simplify the fractions that are serving as exponents.

#(27^(-1/6)*27^(5/6))#

Now apply the second concept mentioned above. Simplify the fractions again.

#(27^(-1/6+5/6))#

#27^(4/6)#

#27^(2/3)#

Use the third concept noted at the top.

#root(3)(27^2)#

Calculate the the value inside the radicand and rewrite.

#root(3)(729)#

Find the cube root by rewriting the radicand again. Follow the fourth concept after.

#root(3)(9^3)#

#9#