If the equation e^≤ft||x| – 2| + b = 2 has four solution then b lies…

$e^{\left||x| – 2\right| + b} = 2$

so $\left||x| – 2\right| + b=\log 2$

$\left||x| – 2\right|=\log 2-b$

so for 4 solutions

$0<\log 2-b<2$

so $-2<b-\log 2<0$

$\log|2|-2<b<\log|2|$