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Thread: Log Law Intuition

  1. #1
    Junior Member
    May 2011

    Log Law Intuition

    What is the intuition for this logarithm law?

    logbx = (logax) / (logab)

    There is intuition for other log laws. For example:

    log(100) + log (1000)
    = 2 factors of 10 + 3 factors of 10
    = 5 factors of 10
    = 5
    = log (100x1000)
    = (2 + 3) factors of 10
    = 5 factors of 10
    = 5

    So more generally, log(a) + log(b) = log(ab). We're just counting the factors of 10 in a and b, so it doesn't matter whether we count them separately or together. It's equal either way.

    Is there intuition for the equation in red above? Although it's trivial to prove, seeing the equation (and using it) in isolation seems more mechanical (e.g. plugging numbers into the Pythagorean Theorem because I see a right triangle) than logical (e.g. solving x + 5 = 24 by subtracting 5 from both sides).
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  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    $\displaystyle \displaystyle \begin{align*}x &= b^y \\ \log_b{x} &= y \end{align*}$


    $\displaystyle \displaystyle \begin{align*} x &= b^y \\ \log_a{x} &= \log_a{\left(b^y\right)} \\ \log_a{x} &= y\log_a{b} \\ \frac{\log_a{x}}{\log_a{b}} &= y\end{align*}$

    Therefore $\displaystyle \displaystyle \log_b{x} = \frac{\log_a{x}}{\log_a{b}}$.
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  3. #3
    MHF Contributor

    Apr 2005
    I originally wrote just what Prove It did (only slower!). However, you seem to be using "intuition" to mean something different from a regular proof.

    Perhaps this is the kind of thing you mean:
    $\displaystyle u= log_a(x)$ means that x has u factors of a. $\displaystyle v= log_b(x)$ means x has v factors of b. $\displaystyle w= log_b(a)$ means that a has w factors of b. Well, if x has u factors of a and each a has w factors of b, then x has uw factors of b.

    That is, v= uw so u= v/w which translates as
    $\displaystyle log_a(x)= \frac{log_b(x)}{log_b(a)}$
    Last edited by HallsofIvy; May 27th 2011 at 08:43 AM.
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