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Another log problem

  1. Jul 27, 2007 #1
    Hi, I've been trying to work out:

    loga(2+x) = 2 + logax

    This is as far as I can get:

    loga(2+x) - logax =2
    so: loga((2+x)/x) =2
    loga((2+x)/x) =log100 (I'm not even sure where I could go with this one though).

    Relevant equations include:

    log x +log y = log xy
    log x - log y = log x/y

    I'd really appreciate any hints you can give me. Thanks.
  2. jcsd
  3. Jul 27, 2007 #2
    It looks like you represented the 2 on the right side as a common log of 100, but the left side is log base a, so that isn't very useful in this case. Since log base a ((2+x)/x) = 2, try thinking more in terms of, "a raised to the second power is equal to ((2+x)/x)."
  4. Jul 27, 2007 #3
    I tried to follow this trail of thought, but it didn't seem to take me anywhere:

    a^2 =(2+x)/x

    xa^2=2+x or alternatively a=root((2+x)/x)

    ..but I can't see how any of this will get rid of the a, which is ideally what I'd like to do to solve for x.
  5. Jul 27, 2007 #4


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    You don't want to get rid of the a. You want to solve for x in terms of a. Got that? Solve for x, not a.
  6. Jul 27, 2007 #5
    ohhh I see! I've been quite thick really, of course I don't want to get rid of the a.

    Ok so: a[tex]^{2}[/tex]=2+x/x
    x(a[tex]^{2}[/tex]-1) =2
    x= 2/(a[tex]^{2}[/tex]-1)

    Is this right? If yes, can it be simpler?
  7. Jul 27, 2007 #6


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    It looks plenty simple to me.
  8. Jul 27, 2007 #7
    Thanks for your help both of you :)
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