Pointing out the negativity....

Here's a more effecient method for when there are more moduli in the equation.

1.Equate the terms in the modulus to 0 and find the value of the variable. These will be the critical points. In this case, putting $x=0$ and $2^{x-1}-1=0$ The critical points will be 0 and 1.

2.Take all the possible cases. Substitute random values in the modulus according to the case and check whether the value of the whole modulus is positive or negative. If it is negative change the sign else don't change.For eg, when $x<0$ $2^{|x|}$ will become $2^{-x}$ and in $|2^{x-1}-1|$ Substituting x = -1, we get -0.75 and hence the sign should be changed. So for this case the final equation will become $2^{-x}+2^{x-1}-1=2^{x-1}+1$ Solving for x, we get $x=\boxed{-1}$ which lies in the range we took in the case, i.e. x<0.

3.Similarly for the cases $0<=x<1$ and $x>=1$ we get $x=1$ and $x>=1$ But the 2nd case does not include 1 so ONLY FOR THAT CASE x =1 does'nt satisfy. So the complete answer is $x={-1} U [1,infinity)$