The solution set of the equation

mod(2^x-1)+mod(4-2^x)≤3 is,

For x<=0,

1-2^x +4-2^x<=3

2<=2^(x+1)

x+1>=1

x>=0   .. so x=0 belongs to solution.

0<x<=2

2^x-1+4-2^x>=3

3>=3 -------- always true so [0,2]  is a solution now

x>2

2^x-1+2^x-4<=3

2^(x+1)<=8=2^3

x+1<=3

x<=2 ---------no solution in this case

so the final solution is x [0,2]