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Precalculus logarithm question

  1. Jul 30, 2004 #1
    I'm trying to use LOGs other than log base 10 and base e on my TI-86. Can I accomplish this like this?:

    log base a of b = (log base 10 of b) / (log base 10 of a) ?

    or is it:

    log base a of b = (ln b) / (ln a) ?

    Help needed ASAP. Thanks!
     
  2. jcsd
  3. Jul 30, 2004 #2
    Both are correct.
     
  4. Jul 30, 2004 #3

    Galileo

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    I'll use [itex]^a\log b[/itex] for the log base a of b.

    For any positive a,b:
    [tex]^a\log b=x \iff a^x=b \iff \ln a^x=\ln b \iff[/tex]
    [tex] x\ln a = \ln b \iff x=\frac{\ln b}{\ln a}[/tex]

    I took the natural logarithm, but that was a complety arbitrary.
    Therefore:

    [tex]^a\log b=\frac{\ln b}{\ln a}=\frac{^{10}\log b}{^{10}\log a}=\frac{^y\log b}{^y\log a}[/tex]
    for any base y.
     
    Last edited: Jul 30, 2004
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