Please give answer as soon as possible........

Solve for x

2^|x+1|+2^|x|=6

the eqn becomes 2^-(x+1)+2^(-x)=6

write it as 2^(-x)+2*2^(-x)= 12

or 3*2^(-x)= 12

or 2^(-x)= 4

or x= – 2

case 2: when x belongs to (-1, 0]

the eqn becomes 2^(x+1)+2^(-x)=6

putting y= 2^x we get

2y^2+1= 6y

or 2y^2 – 6y+1=0

or y= (3 ± sqrt(7))/2

but x lies in (-1, 0], so 2^x=y must be lie in (½, 1]. obviously (3 + sqrt(7))/2 is greater than 1 while (3 – sqrt(7))/2 is less than ½. so both (3 – sqrt(7))/2 and (3+sqrt(7))/2 are rejected. so there is no solution in this interval.

case 3: when x belongs to (0, inf)

2^(x+1)+2^x= 6

2^x(2+1)= 6

or 2^x= 2

or x= 1

so the final soln is x= – 2, 1