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Math Help - Algebra problem with unknown value on both sides

  1. #1
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    Algebra problem with unknown value on both sides

    Hi all,

    I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

    256^n = n

    I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

    Thanks in advance for any help any one can give me.
    Neil
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  2. #2
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    Quote Originally Posted by nastev View Post
    Hi all,

    I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

    256^n = n

    I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

    Thanks in advance for any help any one can give me.
    Neil
    Isn't it true that y = n and b = 256? If you're talking about natural logarithms, the final step would be to apply e to both sides of the equation.
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  3. #3
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    Thanks for the reply. I'm still not understanding how apply e will help. (though I may be missing something obvious since my algebra is pretty rusty)

    If I apply the log formula I get n = log256(n) (where log256 means log base 256). I don't understand where I go from here to derive n though. Possibly I might be taking the wrong approach to solving this by trying to rearrange it with logs?
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  4. #4
    Bar0n janvdl's Avatar
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    Quote Originally Posted by nastev View Post
    Hi all,

    I'd really appreciate it if some one could show me what steps are required to solve the following equation with the ^ symbol representing power. I get the feeling it should be very simple but I don't seem to be able to work it out:

    256^n = n

    I've tried to rearrange the equation using the and rules but don't seem to be getting any where using this approach.

    Thanks in advance for any help any one can give me.
    Neil
    First of all you need only look at n where n > 0, since f(n) < 0 isn't really important here.

    Then look at the growth rates of f(n) = n and f(n) = 256^n.

    The latter grows at an astonishing rate, while the other grows gradually, moving equally along the x and y axis. Wherever f(n) = n is, f(n) = 256^n will always be much further above it on the graph.
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  5. #5
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    Just logarithms aren't going to help you. Here's what I would do:

    From 256^n= n, divide both sides by 256^n to get 1= n\left(\frac{1}{256}\right)^n. Since a^b= e^{ln(a^b)}= e^{b ln(a)}, we can write \left(\frac{1}{256}\right)^n= e^{n (-ln(256))} so n\left(\frac{1}{256}\right)^n= ne^{n(-ln(256))}= 1

    Now multiply both sides by -ln(256):
    \left(n(-ln(256))e^{n(-ln(256))}= -ln(256)

    Let x= n(-ln(256)) and that becomes xe^x= -ln(256).

    That cannot be solved in terms of "elementary functions" but It can be solved using "Lambert's W function" (also called the "omega function" and the "productlog function") which is specifically defined as the inverse function to xe^x. Applying that to both sides of this equation,
    W(xe^x)= x= W(-ln(256)).

    Since x= n(-ln(256)), n= -\frac{x}{ln(256)} so that n= -\frac{W(-ln(256))}{ln(256)}.

    The W function can be expanded in the MacLaurin series
    \sum_{k=1}^\infty \frac{(-1)^{k-1}k^{k-2}}{(k-1)!}x^k
    Last edited by HallsofIvy; December 18th 2010 at 12:35 PM.
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