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Math Help - how to solve log2 (x) = log2(2x-1)

  1. #1
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    how to solve log2 (x) = log2(2x-1)

    Hi

    I tried:
    log2(x) - log2(2x-1) = 0

    but then log2(0) not possible so could go nowhere with that.

    I know 2^(log2(2x-1)) = 2^log2(x) but does that help me?

    Bit stuck on this one, please help.

    Angus
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  2. #2
    Senior Member BAdhi's Avatar
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    Re: how to solve log2 (x) = log2(2x-1)

    I think what you've stated is \log_2(x)-\log_2(2x-1)=0

    if so it can be adjusted as,

    \log_2(x)=\log_2(2x-1)

    remove the logs and proceed
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: how to solve log2 (x) = log2(2x-1)

    Writing \log_{2} x = \log_{2} (2 x -1) and writing x= 2x-1 is exactly the same...

    Kind regards

    \chi \sigma
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  4. #4
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    Re: how to solve log2 (x) = log2(2x-1)

    Quote Originally Posted by angypangy View Post
    Hi

    I tried:
    log2(x) - log2(2x-1) = 0

    but then log2(0) not possible so could go nowhere with that.
    I don't see where "log2(0)" has anything to do with that. What you have is equivalent to
    log_2\left(\frac{x}{2x-1}\right)= 0
    and, while it is impossible to take the logarithm of 0, the logarithm, to any base, of 1 is equal to 0.
    \frac{x}{2x- 1}= 1 which, of course, is the same as x= 2x- 1.

    I know 2^(log2(2x-1)) = 2^log2(x) but does that help me?
    Well, yes! Since 2^{log_2(a)}= a ( log_2(x) and 2^x are inverse functions) that tells you immediately that 2x- 1= x.

    Or, you could argue that since log_2(x) is a one to one function, log_2(a)= log_2(b) gives, immediately, a= b. That is, because log_2(x)= log_2(2x-1), you must have x= 2x- 1.

    Bit stuck on this one, please help.

    Angus
    Last edited by HallsofIvy; June 22nd 2011 at 12:51 PM.
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  5. #5
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    Re: how to solve log2 (x) = log2(2x-1)

    Loga(x) + Loga(y) = x+y, based on this rule, x= 2x-1. -x=-1, hence x=1.
    substitute to check....
    log2(1)= 0 & log2(2-1)= log2(1)=0

    yip its correct :-)
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