I need to solve (2ar^(a-1))-1 =0 
Answer given in text is r = (2a)^(1/(1-a)), but I get to r =
(2a)^(-1/(1-a))- where am I going wrong?

Is it "but I get to r = (2a)^(-1/(1-a))- where am I going wrong?"   Or

r = (2a)^(-1/(1-a))  ?


if it is:
r = (2a)^(-1/(1-a))-  

Do the two minus signs cancel each other out?  or is one a typo?

Do you get this for your second step

2ar^(a-1)=+1 ?

Later, are you raising both sides to the (1/(a-1)) power?

If these don't help, you might try posting your steps to make it
easier for someone to pinpoint the problem.

Did you convert 1/2a to (2a)^-1?

Are you getting r=(2a)^(-1/(1-a)) or 
r = (2a)^(-1/(a-1))
  = (2a)^(1/(-(a-1)) 
  = (2a)^(1/(-a+1) 
  = (2a)^(1/(1-a))

If you are getting r = (2a)^(-1/1-a) then your problem is after you get 
r^(a-1) = 2a^-1 you are trying to take the (1-a)th root of each side
instead of the (a-1)th leading to
r^[(a-1)/(1-a)] = (2a)^(-1/(1-a)) which clearly doesn't solve for r.

Hope this helps.

Boy, ga picked a horrible default font for trying to communicate math problems!

Instant headache font I think it is called.