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Different ways of expressing logarithms

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]^{2}log(e)=\frac{1}{ln2}[/tex]

    2. Relevant equations
    [tex]^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x)[/tex][tex]^{a}log(x)=\frac{^{b}log(x)}{^{b}log(a)}[/tex]
    3. The attempt at a solution
    How can this be shown? I assume it can be done just using logarithm laws, but I don't see how. I tried manipulating around the factors, but I can't get to it. Can someone please show some way to show what the question asks.
     
  2. jcsd
  3. Jul 26, 2014 #2
    I think you wrongly formatted [itex]log_a(x)[/itex] as [itex]a log(x)[/itex] instead.

    You can use the second relevant equation you have noted above. What can you use as the value of 'b' in that equation to get something that could look like the RHS of your desired proof ? And what is [itex]log_k(k)[/itex] for any k in a valid domain?
     
  4. Jul 26, 2014 #3
    Where?

    Actually I shouldn't have written that equation. I want to know how to do it, without using this equation. The point of the exercise is actually to make an example which leads to this equation. Is this possible?
     
  5. Jul 26, 2014 #4

    Ray Vickson

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    The standard notation for the base-a logarithm of b is ##\log_a b## or ##\log_a(b)##, not ##{}^a \log b## or ##{}^a \log(b)##. Anyway, ##\log_a b = x## means that ##a^x = b##, so you should be able get everything you need from that.
     
  6. Jul 26, 2014 #5
    You could prove the relevant equation, or the question itself.

    Let [itex]log_b(x) = p[/itex], and [itex]log_a(x) = q[/itex]
    Substituting for x in the second equation, by using the definition of logarithm on the first, you should be able to prove the relevant equation.
     
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