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Math Help - Explain why these steps were taken...

  1. #1
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    Explain why these steps were taken...

    There is something which I don't quite understand, and I'm hoping someone could explain it to me.

    Basically, I'm trying to find a statement that expresses this:
    <br />
\log_{ab}x<br />
    in terms of p and q

    <br />
Let,<br />
    <br />
\log_{a}x=p<br />
    <br />
\log_{b}x=q<br />

    <br />
\log_{a}x= \frac{\log_{ab}x}{\log_{ab}a}=p<br />
    <br />
\log_{b}x= \frac{\log_{ab}x}{\log_{ab}b}=q<br />

    I don't understand where the base ab came from in the above two expressions. Is there a way to do it without the ab?
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  2. #2
    Super Member Showcase_22's Avatar
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    I don't understand where the base ab came from in the above two expressions
    This is from the rules of changing the base of logs.

    CHANGING THE BASE OF A LOGARITHM

    Is there a way to do it without the ab?
    It depends on how you define a and b. In the way they are currently represented, they can only be expressed using \log_{ab}.
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  3. #3
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    But then, wouldn't it be:

    <br />
\log_{a}x=\frac{\log_x}{\log_a}<br />

    not

    <br />
\log_{a}x=\frac{\log_{ab}x}{\log_{ab}a}<br />
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  4. #4
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    Hello, maybealways!

    Let: . \begin{array}{c}\log_ax \:=\:p \\ \log_bx \:=\:q \end{array}

    Express \log_{ab}x in terms of p and q.

    \log_{a}x\:=\: \frac{\log_{ab}x}{\log_{ab}a}\:=\:p

    \log_{b}x\:=\: \frac{\log_{ab}x}{\log_{ab}b}\:=\:q

    I don't understand where the base ab came from.
    Is there a way to do it without the ab?

    Yes, but it requires some Olympic-level gymnastics.

    We have: . \log_ax \:=\:p \quad\Rightarrow\quad x \:=\:a^p \quad\Rightarrow\quad \:x^{\frac{1}{p}} \:=\:a .[1]

    . . . .and: . \log_bx \:=\:q \quad\Rightarrow\quad x \:=\:b^q \quad\Rightarrow\quad x^{\frac{1}{q}}\:=\:b .[2]


    Mutliply [1] and [2]: . x^{\frac{1}{p}}\cdot x^{\frac{1}{q}} \:=\:a\cdot b \quad\Rightarrow\quad x^{\frac{1}{p}+\frac{1}{q}}\:=\:ab


    \text{Take logs (base }ab)\!:\;\;\log_{ab}\left(x^{\frac{1}{p} + \frac{1}{q}}\right) \:=\;\underbrace{\log_{ab}(ab)}_{\text{This is 1}}

    . . Then: . \left(\tfrac{1}{p} + \tfrac{1}{q}\right)\log_{ab}(x) \:=\:1 \quad\Rightarrow\quad \log_{ab}(x) \:=\:\frac{1}{\frac{1}{p} + \frac{1}{q}}


    Therefore: . \log_{ab}(x) \;=\;\frac{pq}{p+q}

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