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    Find the range of  x values such that the inequality
     \displaystyle  \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
    holds.

    It's a bit of an awkward question.
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    (Original post by Ano123)
    Find the range of  x values such that the inequality
     \displaystyle  \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
    holds.

    It's a bit of an awkward question.
    Pretty straight forward if you draw a graph. What about it?
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    (Original post by RDKGames)
    Pretty straight forward if you draw a graph. What about it?
    What's your sketch?
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    (Original post by Ano123)
    What's your sketch?
    draw the graph
    find the solution(s)
    look at the range between/out of these solution(s) using the graph and bam u got yourself an answer :yy:
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    (Original post by Ano123)
    What's your sketch?
    Draw both sides as separate functions on the same axis
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    (Original post by Ano123)
    What's your sketch?
    Something like this.

    Name:  ImageUploadedByStudent Room1473942846.250370.jpg
Views: 39
Size:  150.9 KB


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    (Original post by Ano123)
    Find the range of  x values such that the inequality
     \displaystyle  \bigg ||2^x-1|-2 \bigg | \leq \left (\frac{1}{2}\right )^x
    holds.

    It's a bit of an awkward question.
    Consider the various cases. Firstly when x>0, then |2^x - 3| < 2^(-x) so if x>log2(3) then 2^x-3<2^(-x) and if 0<x<log2(3) then 3-2^x<2^(-x). Both of these are disguised quadratics, so you should be able to continue from here.
 
 
 
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