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Math Help - help with logarithm eq.

  1. #1
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    help with logarithm eq.

    Hi , I need help on the following:

    Find a value of x such that :

    x = 4(\lceil log_2 (x)\rceil + 1)

    Thanks
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  2. #2
    MHF Contributor

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    The "ceiling" of log_2(x), and so the right side, changes only when x is a power of two. If x= 2^n, 4(log_2(x)+ 1)= 4(n+1). In order for that to be equal to x= 2^n we must have 2^n= 4(n+1), 2^n/4= 2^{n-2}= n+1.

    You might be able to solve that algebraically using the Lambert W function but I would try a numerical method:
    If n= 1, 2^{-1}= 1/2\ne 1+ 1. If n= 2 we have 2^0= 1< 2+1. If n= 3 we have 2^1= 2< 3+ 1. If n= 4 we have 2^2= 4< 4+1. If n= 5 we have 2^3= 8> 5+ 1. That tells us that there must be a solution to 2^{n- 2}= n+ 1 for n a number between n= 4 and n= 5. Okay, try half way between (just because that is simplest). If n= 4.5, [tex]2^{2.5}= 4\sqrt{2}[/quote] which is just slightly more than 4.5+ 1= 4.6. There must be a solution between n= 4 and 4.5 (and so x between 2^4= 16 and [tex]2^{4.5}= 22.67...
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