Solve `|2^x-1|+| 4-2^x| lt 3`

Explanation

`|2^x-1|+| 4-2^x| lt 3`
`2^x-1 = 0 , if x=0`
`4-2^x=0 , if x=2`
Case (i):
`x lt 0 `
`rArr 1-2^x +4- 2^x lt 3 `
`rArr 2 xx 2 ^x lt 3 `
`2 xx 2 ^x lt 2 rArr 2^x gt 1` not possible when `x lt 0 `
Case (ii) : `0 le x lt 2 `
`rArr 2^x -1 + 4 - 2^x lt 3 ` not possible
Case (iii) `x ge 2`
`rArr 2xx 2 ^x lt 8 `
`rArr 2^x lt 4 `
`rArr x lt 2 ` not possible
Alternative Method
We have `|2^x-1|+|4-2^x| lt 3 `
But `|(2^x-1) + ( 4-2^x)| le |2^x -1 |+ |4-2^x| lt 3 `
or `3 lt 3 `which is not possible
Hence , three is no real solution