SOLUTION: How would I solve (z+2)^-(2/3) = 3 i dont know if this is easy to understand but the first equation is to the negative two thirds power.

Algebra ->  Radicals -> SOLUTION: How would I solve (z+2)^-(2/3) = 3 i dont know if this is easy to understand but the first equation is to the negative two thirds power.      Log On


   

Question 151986: How would I solve
(z+2)^-(2/3) = 3
i dont know if this is easy to understand but the first equation is to the negative two thirds power.
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for x:
%28z%2B2%29%5E%28-2%2F3%29+=+3 Raise both sides to the power of %28-3%2F2%29
%28%28z%2B2%29%5E%28-2%2F3%29%29%5E%28-3%2F2%29+=+3%5E%28-3%2F2%29 To the left side, apply the "product rule for exponents" %28M%5Ea%29%5Eb+=+M%5E%28ab%29
%28z%2B2%29+=+3%5E%28-3%2F2%29
z%2B2+=+%281%29%2F%283%5E%283%2F2%29%29 This is supposed to be: 1over3%5E%283%2F2%29 and 3%5E%283%2F2%29+=+sqrt%2827%29
z%2B2+=+1%2Fsqrt%2827%29 Multiply top and bottom bysqrt%2827%29
z%2B2+=+sqrt%2827%29%2F27 Now subtract 2 from both sides.
z+=+sqrt%2827%29%2F27+-+2
z+=+%28sqrt%2827%29-54%29%2F27