How do you simplify #(49/81) ^ (-1/2)#?

Question

5295 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-03T16:16:23

9/7

#1/sqrt(49/81)# =#1/(7/9)# = #9/7#

Answer 2 · 2018-05-03T16:17:00

#9/7#

#(a/b)^(m/n)=root(n)((a/b)^m)#
If #m/n# is negative
#(a/b)^-(m/n)=1/root(n)((a/b)^m)#

SO:
#(49/81)^(-1/2)=1/sqrt((49/81)^1)=1/(7/9)=9/7#

Answer 3 · 2018-05-03T16:19:00

Simplified we should get #+-9/7#

To solve this we need to remember that #x^{1/2}= sqrt(x)#
Aslo #x^(-1)=1/x#
And lastly we should recognise that
#49=7^2# and #81=9^2#

If we do, it's quite straightforward:
#(49/81)^-(1/2)=(9^2/7^2)^(1/2)=(9/7)^(2*1/2)=9/7#

As #(-sqrt(x))^2=(sqrt(x))^2=x#

the full simplification should then be #+-9/7#