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Thread: Is this logarithm true?

  1. #1
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    Is this logarithm true?

    How would I show that 1/log(subscript a) b = log(subscript b) a ?
    I'm not sure where to start because of the different bases.
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  2. #2
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    Quote Originally Posted by TN17 View Post
    How would I show that 1/logab = logba ?
    I'm not sure where to start because of the different bases.
    Is this the question?
    $\displaystyle \log_a(b)=\dfrac{1}{\log_b(a)}$
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  3. #3
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    Yes, sorry for the typo.
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  4. #4
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    Quote Originally Posted by Plato View Post
    Is this the question?
    $\displaystyle \log_a(b)=\dfrac{1}{\log_b(a)}$
    Yes, that's what I meant. Sorry for the typo.
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  5. #5
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    You should know that $\displaystyle \log_a(b)=\dfrac{\ln(b)}{\ln(a)}$.
    So what does $\displaystyle \dfrac{\ln(a)}{\ln(b)}=~?$
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  6. #6
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    Hm... we haven't learned ln yet.
    The teacher mentioned it really briefly in class, but didn't do a lesson on it.
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  7. #7
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    Quote Originally Posted by TN17 View Post
    Hm... we haven't learned ln yet.
    The teacher mentioned it really briefly in class, but didn't do a lesson on it.
    Well that is the proof.
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  8. #8
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    Couldbn't I say that
    log (subscript b) a = log a / log b is true for any base/subscript.

    Then 1 / [log (subscript a) b] = 1 / [log b / log a]
    Where 1 / [log b / log a]
    =log a / log b

    Now use the fact at the top to change this into

    log a / log b = log (subscript b) a

    Therefore, 1 / [log (subscript a) b] = log (subscript b) a.

    ?
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  9. #9
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    You have the idea.$\displaystyle ~~\log_b(a)=\dfrac{\ln(a)}{\ln(b)}$
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  10. #10
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    Quote Originally Posted by TN17 View Post
    How would I show that 1/log(subscript a) b = log(subscript b) a ?
    I'm not sure where to start because of the different bases.
    Alternatively,

    $\displaystyle log_ab=x\Rightarrow\ b=a^x$

    $\displaystyle log_ba=y\Rightarrow\ a=b^y=\left(a^x\right)^y=a^{xy}\Rightarrow\ xy=1$

    $\displaystyle \Rightarrow\ y=\frac{1}{x}\Rightarrow\ log_ba=\frac{1}{log_ab}$
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  11. #11
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    I understand up to where you wrote a = b(superscript y)
    I don't understand how you got a(superscript x)(superscript y) and onward...
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  12. #12
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    Quote Originally Posted by TN17 View Post
    I understand up to where you wrote a = b(superscript y)
    I don't understand how you got a(superscript x)(superscript y) and onward...
    Your basic problem is not knowing or understanding the basic definitions.

    $\displaystyle \log_b(a)=t\text{ if and only if }a=b^t,~~a>0~\&~b>0.$
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  13. #13
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    Quote Originally Posted by TN17 View Post
    I understand up to where you wrote a = b(superscript y)
    I don't understand how you got a(superscript x)(superscript y) and onward...
    $\displaystyle a=b^y$

    Now substitute in the value we have for $\displaystyle b$, which is $\displaystyle b=a^x$

    $\displaystyle a=b^y=\left(a^x\right)^y$ since $\displaystyle b=a^x$

    Using the laws of indices... $\displaystyle \left(c^p\right)^q=c^{pq}$

    similar to $\displaystyle \left(2^2\right)^3=\left(2^2\right)\left(2^2\right )\left(2^2\right)=2^{2+2+2}=2^6=2^{(2)(3)}$

    Therefore $\displaystyle \left(a^x\right)^y=a^{xy}$

    but this is equal to $\displaystyle a$

    hence $\displaystyle xy=1$

    Although this is fine from the perspective of indices, Plato's method is the better one to get used to.
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  14. #14
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    Hello, TN17!

    $\displaystyle \text{Show that: }\:\dfrac{1}{\log_a(b)} \:=\: \log_b(a)$

    Let the right side equal $\displaystyle \,P\!:\;\;\log_b(a) \:=\:P$ .[1]

    Then we have: .$\displaystyle b^P \:=\:a$

    Take logs, base $\displaystyle a\!:\;\;\log_a(b^P) \:=\:\underbrace{\log_a(a)}_{\text{This is 1}}$

    . . . . . . $\displaystyle P\cdot\log_a(b) \:=\:1 \quad\Rightarrow\quad \log_a(b) \:=\:\dfrac{1}{P}$

    Substitute [1]: .$\displaystyle \log_a(b) \;=\;\dfrac{1}{\log_b(a)} $

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  15. #15
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    Thanks. : )
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