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`        Please give answer as soon as possible........Solve for x 2^|x+1|+2^|x|=6`
10 months ago

```							Hiii There will be two cases for this question , In case 1 – Open x+1 and x as positive quantity  and solve for value of x, In Case -2 – Open it as -(x+1) and -x , and solve for value of x , Take the intersection of both case- 1 and case -2 , that will be your solution , Hope this hep , If does not clear  ask me in the comment section, i will give solutions.
```
10 months ago
```							hello king. vikas’ answer above is a bit incompletecase 1: when x belongs to (-inf, – 1]the eqn becomes 2^-(x+1)+2^(-x)=6write it as 2^(-x)+2*2^(-x)= 12or 3*2^(-x)= 12or 2^(-x)= 4or x= – 2case 2: when x belongs to (-1, 0]the eqn becomes 2^(x+1)+2^(-x)=6putting y= 2^x we get2y^2+1= 6yor 2y^2 – 6y+1=0or y= (3 ± sqrt(7))/2but 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= 62^x(2+1)= 6or 2^x= 2or x= 1so the final soln is x= – 2, 1kindly approve :)
```
10 months ago
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